Let $\zeta_3$ be the third root of unity.
1) Does it hold that: $\mathbb{Q}(\sqrt{2},\zeta_3)=\mathbb{Q}(\sqrt{2}+\zeta_3)$ ?
2) Does it hold that $\mathbb{Q}(\sqrt[3]{2},\zeta_3)=\mathbb{Q}(\sqrt[3]{2}\zeta_3)$?
My attempt for 1) is to compute the minimal polynomial of $\sqrt{2}+\zeta_3$ as $p(x)=x^4+2x^3-x^2-2x+7$ which is of degree 4. But since $|\mathbb{Q}(\sqrt{2},\zeta_3):\mathbb{Q}|=6$ these can't be equal. Is my proof correct?
For the second one I am not sure.
On
1) There is a standard trick for this kind of problem. You have obviously that $\mathbb{Q}(\sqrt{2},\zeta_3)\supset \mathbb{Q}(\sqrt{2} +\zeta_3)$. Now we show the other containment; we have: \begin{equation} \alpha = \sqrt{2} +\zeta_3 \quad \Longrightarrow \quad (\alpha - \sqrt{2})^3 = \zeta_3^3 \end{equation} Using $\zeta_3^3=1$ and doing the computation we have: \begin{equation} \sqrt{2} = \dfrac{\alpha^3+6\alpha-1}{3\alpha^2+2} \end{equation} Then $\sqrt{2}\in \mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt{2} +\zeta_3)$. Having $\sqrt{2}\in \mathbb{Q}(\sqrt{2} +\zeta_3)$ we also have $\zeta_3\in \mathbb{Q}(\sqrt{2} +\zeta_3)$, so $\mathbb{Q}(\sqrt{2} +\zeta_3)\supset \mathbb{Q}(\sqrt{2},\zeta_3)$ and the equality.
2) $[\mathbb{Q}(\sqrt[3]{2}\zeta_3) : \mathbb{Q}] = 3$ since its minimum polynomial of $\sqrt[3]{2}\zeta_3$ is $(x^3-2)$ that is irreducible for Eisenstein criterion. $[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3$ for the same reason above and $[\mathbb{Q}(\zeta_3) : \mathbb{Q}] = 2$ because its minimum polynomial is $(x^2+x+1)$. Since $2$ and $3$ are coprime $[\mathbb{Q}(\sqrt[3]{2},\zeta_3) : \mathbb{Q}] = 6$. So $\mathbb{Q}(\sqrt[3]{2},\zeta_3) \neq \mathbb{Q}(\sqrt[3]{2}\zeta_3)$.
$\zeta_3$ is a root of $x^2 + x + 1$ since $x^3 - 1 = (x - 1)(x^2 + x + 1)$. So its degree is $2$ not $3$.
For 2) you can do the same kind of thing. The minimal polynomial of $\sqrt[3]2 \zeta_3$ is $x^3 - 2$ so $\mathbb{Q}(\sqrt[3]2 \zeta_3) \ne \mathbb{Q}(\sqrt[3]2, \zeta_3)$. It's a 3-dimensional subspace of a 6-dimensional $\mathbb{Q}$-vector space.