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 $4$.
Here is an extremely easy way of showing that $\sqrt{2} \in \Bbb{Q}(\sqrt{2} + \sqrt[3]{2})$. Write $\alpha = \sqrt{2} + \sqrt[3]{2}$. Then $(\alpha-\sqrt{2})^3 = 2$, so that
$$\alpha^3 - 3\alpha^2(\sqrt{2}) + 6\alpha -2\sqrt{2} = 2.$$
It follows that $\alpha^3+ 6\alpha - 2 = \sqrt{2}(3\alpha^2+ 2)$, so that
$$\sqrt{2} = \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}.$$
Since $\Bbb{Q}(\alpha)$ is a field, the expression on the right hand side is in here, so that $\sqrt{2}$ is in here. No using horrible row reduction to calculate anything! It follows that $\sqrt[3]{2}$ is in here. Hence $\Bbb{Q}(\alpha)$ contains the fields $\Bbb{Q}(\sqrt{2})$ and $\Bbb{Q}(\sqrt[3]{2})$ which means that $[\Bbb{Q}(\alpha) : \Bbb{Q}]$ is a multiple of 3 and 2. Since we already know that it is at most $6$, it follows now that since $2$ and $3$ are coprime that $[\Bbb{Q}(\alpha) : \Bbb{Q}] = 6$. Whatever monic polynomial of degree 6 that you find for $\alpha$ over $\Bbb{Q}$ will then be irreducible, and hence will be the minimal polynomial of $x$ over $\Bbb{Q}$.
Best Answer
Your first minimum polynomial is correct and your method for showing containment is fine. In regards to finding the minimum polynomial over $\mathbb Q(\sqrt{3})$ the usual method is to write the equation
$$x=\sqrt{1+\sqrt{3}}$$
in terms of elements in the base field. Squaring both sides we have
$$x^2=1+\sqrt{3}$$
notice that $1+\sqrt{3} \in \mathbb Q(\sqrt{3})$. In order to show this polynomial is irreducible you have to be a little clever. Notice that $[\mathbb Q(\sqrt{3}):\mathbb Q]=2$ and $[\mathbb Q(\alpha) : \mathbb Q]=4$ so that tells us the degree of the minimum polynomial of $\alpha$ over $\mathbb Q(\sqrt{3})$ is either $2$ or $4$. Hence our polynomial is irreducible because it cannot be of a lower degree.