Let us perform tuple arithmetic to obtain \(t\) in the simplest possible form: \[t = \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}+ 2\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}+ \begin{bmatrix} 2 \\ -2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}. \] Now, \(t_1\) refers to the first entry of \(t\) and \(t_2\) refers to the second entry of \(t\). Hence, \(t_1+t_2^2 = 2+(-1)^2 = 3\).

Note that if \(n\in\mathbb{N}\), then \(n-2\) is an integer. So the only possibilities for \((n-2)^2\) to be less than or equal to \(10\) are \(0,1,4,9\). Thus, the possibilities for \(n-2\) are \(0, \pm 1, \pm 2, \pm 3\). As \(n\) must be a natural number, the elements of the set are just \(1, 2, 3, 4, 5\).

We consider two cases based on the value of \(b\).

Case 1. \(b \geq 0\). Since \(a \geq 0\), \(|a+b|=1\) can be simplified as \(a+b=1\). Because \(a\) and \(b\) are required to be nonnegative integers, we must have \[a = 0, b = 1,\] or \[a = 1, b = 0.\]

Case 2. \(b = -1\). Hence, we need \(a\) to satisfy \(|a-1|=1\). The only possibilities are \(a = 0\) and \(a = 2\).

Combining the two cases, we see that all the elements of the given set are \(\begin{bmatrix} 0\\ 1\end{bmatrix}\), \(\begin{bmatrix} 1\\ 0\end{bmatrix}\), \(\begin{bmatrix} 0\\ -1\end{bmatrix}\), and \(\begin{bmatrix} 2\\ -1\end{bmatrix}\).

Recall that over \(GF(2)\), \(1+1 = 0\). Hence, \begin{eqnarray*} 1 + (1 + (1 + 1)\cdot 1) & = & 1 + (1 + 0 \cdot 1) \\ & = & 1 + (1 + 0) \\ & = & 1 + 1 \\ & = & 0. \end{eqnarray*}

Note that \(\sqrt{6.25} = 2.5 = \frac{5}{2}\) which is rational. It is well-known that \(\pi\) is irrational and the square root of any prime number is irrational. Hence, \(A\cap \mathbb{Q} = \{-5, \sqrt{6.25}\}\) and \(A \setminus B =\{-5,\sqrt{17}\}\) since \(-5\) and \(\sqrt{17}\) are the only elements in \(A\) that are not in \(B\).

Recall that the multiplicative inverse of \(4-\sqrt{5}\) is a number \(\alpha\) such that \(\alpha (4- \sqrt{5}) = 1\). In this question, we need to give \(\alpha\) in the form \(a + b\sqrt{5}\) with \(a,b\in \mathbb{Q}\).

Hence, we want to find \(a,b\in \mathbb{Q}\) such that \((a+b\sqrt{5})(4-\sqrt{5}) = 1\).

Expanding the left-hand side gives: \begin{eqnarray*} \\ (a+b\sqrt{5})(4-\sqrt{5}) & = & a(4-\sqrt{5}) +b\sqrt{5}(4-\sqrt{5}) \\ & = & 4a-a\sqrt{5} +4b\sqrt{5}-5b \\ & = & (4a-5b) + (-a+4b)\sqrt{5}. \end{eqnarray*} So, we need \((4a-5b) + (-a+4b)\sqrt{5}\) to be equal to \(1\). Because \(\sqrt{5}\) is irrational and \(-a+4b\) is rational, we must have \(-a+4b = 0\). This means that we need \(4a-5b=1\).

Now, \(-a + 4b = 0\) means \(a = 4b\). Substituting this into \(4a-5b=1\), we get \(4(4b)-5b=1\), implying that \(11b = 1\). So, \(b = \frac{1}{11}\). Hence, \(a = \frac{4}{11}\). Therefore, the answer is \(\frac{4}{11}+\frac{1}{11}\sqrt{5}\).

(One can in fact arrive at the answer much more quickly as follows: In effect, we want to clear the radical in the denominator of \(\frac{1}{4-\sqrt{5}}\). To do so, we simply compute \(\frac{1}{4-\sqrt{5}}\cdot \frac{4+\sqrt{5}}{4+\sqrt{5}} = \frac{4+\sqrt{5}}{4^2 - (\sqrt{5})^2} = \frac{4+\sqrt{5}}{16 - 5} = \frac{4+\sqrt{5}}{11}.\) Here, \(4 + \sqrt{5}\) is called the conjugate of \(4-\sqrt{5}\).)

Note that \(x^2 \geq 0\) and \(3 < \pi < 4\). So the smallest positive integer in the range of \(f\) is \(4\) and is attained at \(x = \sqrt{4 - \pi}\).

We need to show that for every \(\begin{bmatrix} a\\ b\end{bmatrix} \in \mathbb{R}^2\), there exist real numbers \(x\) and \(y\) such that \(T\left( \begin{bmatrix} x \\ y \end{bmatrix} \right)=\begin{bmatrix} a \\ b \end{bmatrix}\).

It suffices to show that the following system always has a solution: \begin{eqnarray*} 2x - y & = & a \\ x + y & = & b \end{eqnarray*} Adding the two equations gives \(3x = a+b\), implying that \(x = \frac{a+b}{3}\). From the second equation, we get that \(y = b - x = \frac{-a + 2b}{3}\).

Thus, \(T\left( \begin{bmatrix} \frac{a+b}{3} \\ \frac{-a+2b}{3} \end{bmatrix} \right)=\begin{bmatrix} a \\ b \end{bmatrix}\) for all \(a,b\in\mathbb{R}\).

Since \(f(ax^2 + bx + c) = \begin{bmatrix} a-2b+c \\ b- c\end{bmatrix},\) we see that \( f(ax^2 + bx + c) = \begin{bmatrix} 1 \\ 0\end{bmatrix}\) if and only if \[ \begin{bmatrix} a-2b+c\\b-c\end{bmatrix} = \begin{bmatrix} 1 \\ 0\end{bmatrix}.\]

Comparing the corresponding entries on both sides, we get that \(a-2b+c = 1\) and \(b-c=0\). Now, \(b-c = 0\) implies that \(b = c\). Thus, \(a-2b+c = 1\) implies that \(a = 1+2b-c = 1+c\). It follows that \(f(u) = \begin{bmatrix} 1 \\ 0\end{bmatrix}\) if and only if \(u = (1+c)x^2 + cx + c\) where \(c \in \mathbb{R}\).

First, we need to check that \(\mathbb{Q}(\sqrt{2})\) is closed under addition and multiplication; that is, adding two numbers in \(\mathbb{Q}(\sqrt{2})\) results in another number in \(\mathbb{Q}(\sqrt{2})\) and multiplying two numbers in \(\mathbb{Q}(\sqrt{2})\) also results in another number in \(\mathbb{Q}(\sqrt{2})\).

For addition, note that \( (a+b\sqrt{2}) + (c + d\sqrt{2}) = (a+c) + (b+d)\sqrt{2}\). If \(a,b,c,d\) are rational, so are \(a+c\) and \(b+d\).

For multiplication, note that \( (a+b\sqrt{2}) \cdot (c + d\sqrt{2}) = (ac+2bd) + (ad+bc)\sqrt{2}\). If \(a,b,c,d\) are rational, so are \(ac+2bd\) and \(ad+bc\).

Hence, adding or multiplying numbers in \(\mathbb{Q}(\sqrt{2})\) gives numbers that belong to the set.

Next, we need to check that properties of a field are satisfied:

1. For all \(a, b,\) and \(c \in \mathbb{F}\), the following hold: \(a + (b+c) = (a+b)+ c\) and \(a\cdot (b\cdot c) = (a\cdot b)\cdot c\). (Associativity of addition and multiplication.)

2. For all \(a\) and \(b \in \mathbb{F}\), the following hold: \(a + b = b + a\) and \(a\cdot b = b\cdot a\). (Commutativity of addition and multiplication.)

3. For all \(a, b,\) and \(c \in \mathbb{F}\), the following holds: \(a \cdot (b+c) = a \cdot b+ a\cdot c\). (Left-distributivity.)

4. There exists an element of \(\mathbb{F}\), denoted by 0, such that for all \(a \in \mathbb{F}, a + 0 = a\). (Existence of additive identity.)

5. There exists an element of \(\mathbb{F}\) not equal to 0, denoted by 1, such that for all \(a \in \mathbb{F}, a \cdot 1 = a\). (Existence of multiplicative identity.)

6. For each \(a \in \mathbb{F}\), there exists an element \(-a \in \mathbb{F}\), such that \(a+(-a) = 0\). (Existence of additive inverse.)

7. For each \(a \in \mathbb{F}\) not equal to 0, there exists an element \(a^{-1} \in \mathbb{F}\), such that \(a\cdot a^{-1} = 1\). (Existence of multiplicative inverse.)

It seems like a monumental task to check all the listed properties. However, \(\mathbb{Q}(\sqrt{2})\) is a subset of the set of real numbers. If we use the fact that the real numbers form a field, then properties 1 through 5 come for free.

So we need to check properties 6 and 7. The additive inverse of \(a + b\sqrt{2}\) is simply \( (-a) + (-b)\sqrt{2} \). As \(-a\) and \(-b\) are rational numbers, \( (-a) + (-b)\sqrt{2} \) is an element of \(\mathbb{Q}(\sqrt{2}\). Hence, additive inverses exist for all elements of \(\mathbb{Q}(\sqrt{2}\).

We now find the multiplicative inverse of \(a + b\sqrt{2}\) when \(a + b \sqrt{2} \neq 0\).

Note that not both \(a\) and \(b\) can be zero. Using the irrationality of \(\sqrt{2}\), we see that \(a - b\sqrt{2} \neq 0\). Then \begin{eqnarray*} \frac{1}{a + b \sqrt{2}} & = & \frac{1}{a + b \sqrt{2}} \cdot \frac{a-b\sqrt{2}}{a - b \sqrt{2}} \\ & = & \frac{a-b\sqrt{2}}{a^2 - 2b^2} \\ & = & \frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2} \end{eqnarray*} As \(\frac{a}{a^2-2b^2}\) and \(\frac{b}{a^2-2b^2}\) are rational numbers, \(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\) is the multiplicative inverse of \(a + b\sqrt{2}\) from \(\mathbb{Q}(\sqrt{2})\).

Since \(a \neq 0\), we know that \(a^{-1}\) exists and that \(a^{-1} a=1\). Let \(b = \left(a^{-1}\right)^k\). Then \(ba^k = \left(a^{-1}\right)^k a^k = \left(a^{-1}a\right)^k = 1^k = 1\). Hence, by definition, \(b\) is the multiplicative inverse of \(a\).