In general, you can find the minimum polynomial for an algebraic number $\alpha$ by determining the smallest power of $\alpha$ for which $\{1,\alpha,\alpha^2,\ldots,\alpha^n\}$ is linearly dependent over $\mathbb{Q}$.
For example, let $\alpha=\sqrt{2} + \sqrt{3}$. Then
\begin{align*}
\alpha^2 &= 5+2\sqrt{6} \\\
\alpha^3 &= 11\sqrt{2}+9\sqrt{3} \\\
\alpha^4 &= 49+20\sqrt{6}
\end{align*}
Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ are linearly independent over $\mathbb{Q}$, we can think of the powers of $\alpha$ as vectors:
$$
1 = \begin{bmatrix}1\\\ 0\\\ 0\\\ 0\end{bmatrix},
\qquad
\alpha = \begin{bmatrix}0\\\ 1\\\ 1\\\ 0\end{bmatrix},
\qquad
\alpha^2 = \begin{bmatrix}5\\\ 0\\\ 0\\\ 2\end{bmatrix},
\qquad
\alpha^3 = \begin{bmatrix}0\\\ 11\\\ 9\\\ 0\end{bmatrix},
\qquad
\alpha^4 = \begin{bmatrix}49\\\ 0\\\ 0\\\ 20\end{bmatrix}
$$
As you can see, $\{1,\alpha,\alpha^2,\alpha^3\}$ is linearly independent, so $\alpha$ is not the root of any cubic polynomial. However, $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ is linearly dependent, with
$$
\alpha^4 - 10\alpha^2 + 1 \;=\; 0
$$
It follows that $x^4-10x^2+1$ is the minimum polynomial for $\alpha$.
This technique depends on being able to recognize a useful set of linearly independent elements like $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$. Depending on how much Galois theory you know, it may be hard to prove that elements like this are linearly independent over $\mathbb{Q}$. In this case, you can use this technique to guess the minimum polynomial, and then prove that you are correct by proving that the polynomial you found is irreducible.
For example, if $\alpha = i\sqrt[4]{2}$, then
$$
\alpha^2 = -\sqrt{2},\qquad \alpha^3 = -i\sqrt[4]{8},\qquad \alpha^4 = 2.
$$
It seems clear that $ \{1,\alpha,\alpha^2,\alpha^3\} $ is linearly independent, while $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ satisfies $\alpha^4-2 = 0$. Thus $x^4-2$ should indeed be the minimum polynomial over $ \mathbb{Q} $, though the easiest way to prove it is to show that $x^4 -2$ is irreducible. This can be done using Eisenstein's criterion, or by checking that $ x^4-2 $ has no roots and does not factor into quadratics modulo $4$.
Suppose $f$ is an irreducible polynomial in $\Bbb Q[X]$, and it has a nontrivial irreducible factor $g$ over $K = \Bbb Q(\sqrt{-2})$.
$g$ can't have rational coefficients, so its conjugate $\overline g$ is another factor of $f$ over $K$. Since $g$ and $\overline g$ are coprime, $g \overline g$ is also a factor of $f$, but it has rational coefficients, so you must have $f = \lambda g \overline g$, and the degree of $f$ must be even.
In your case, $f$ has degree $3$ so this is impossible : it has to stay irreducible over $K$.
Best Answer
Here's a little help with part c). Can you see why it's enough to show that $i\root3\of2$ is in ${\bf Q}(i,\root3\of2)$ and that both $i$ and $\root3\of2$ are in ${\bf Q}(i\root3\of2)$? Can you see that the first of these inclusions is trivial? Can you see why $(i\root3\of2)^4$ is in ${\bf Q}(i\root3\of2)$, and how that helps you to show that both $i$ and $\root3\of2$ are in ${\bf Q}(i\root3\of2)$?
Actually, maybe I should start with this: do you know what ${\bf Q}(\root3\of2)$ means? If you don't know that, that's the place to start.