I think that I have a problem in the redaction. I have to compute
1) $G=\text{Gal}(\mathbb Q(\sqrt 3,\sqrt 2)/\mathbb Q)$
I know that $\mathbb Q(\sqrt 3,\sqrt 2)$ is the splitting field of $X^4-10X^2+1$, and that the roots of $X^4-10X+1$ are given by $$\{\sqrt 3+\sqrt 2; \sqrt 3-\sqrt 2;\sqrt 2-\sqrt 3;-\sqrt 2-\sqrt 3\}$$
Then I would define $$\sigma _1: \sqrt 3+\sqrt 2\mapsto \sqrt 3-\sqrt 2$$ $$\sigma _2: \sqrt 3+\sqrt 2\mapsto \sqrt 2-\sqrt 3$$ $$\sigma _3: \sqrt 3+\sqrt 2\mapsto -\sqrt 2-\sqrt 3$$ $$\sigma _4: \sqrt 3-\sqrt 2\mapsto \sqrt 2-\sqrt 3$$ $$\sigma _5: \sqrt 3-\sqrt 2\mapsto -\sqrt 2-\sqrt 3$$ $$\sigma _6: \sqrt 2-\sqrt 3\mapsto -\sqrt 2-\sqrt3$$
I remark that $\sigma_3=\sigma_4$, $\sigma _1=\sigma _6$ and $\sigma _2=\sigma _6$ therefore I can eliminate $\sigma _4,\sigma _5$ and $\sigma _6$. Therefore $G=\{1,\sigma _1,\sigma _2,\sigma _3\}\cong\mathbb Z/2\mathbb Z\times \mathbb Z/2\mathbb Z$ because all element are of order $2$. Is my resolution correct ?
2) $G=\text{Gal}(\mathbb Q(i,\sqrt[4]2)/\mathbb Q)$
I know that $\mathbb Q(i,\sqrt[4]2)$ is the splitting field of $X^4-2$ and that the root of $X^2-2$ are the elements of $\{\sqrt[4]2,-\sqrt[4]2,i\sqrt[4]2,-i\sqrt[4]2\}$. But now, I don't know how to continue... I know that the answer is $D_8$, but I have problem to represent it.
3) $G=\text{Gal}(\mathbb Q(\sqrt[4] 5)/\mathbb Q)$
I know that $X^4-5$ is the minimal polynomial of $\sqrt[4]5$ and that only the solution $\pm\sqrt[4]5$ are in $\mathbb Q(\sqrt[4]5)$. But the problem is that I only find $id$ and the automorphism $\sigma :\sqrt[4]5\mapsto -\sqrt[4]5$ but we should have $4$ automorphism, not 2 because $[\mathbb Q(\sqrt[4]5):\mathbb Q]=4$ therefore I would need some help.
Thanks,