Computing the Galois group of $x^4+ax^2+b \in \mathbb{Q}[x] $ [duplicate]

For i) and ii), let the conjugates of $\alpha$ be $\gamma$, $\delta$, and $\epsilon$. Consider the automorphism that maps $\alpha$ to $\gamma$. If it maps $\gamma$ to $\delta$ (and then, as you can show, $\delta$ to $\epsilon$, and $\epsilon$ to $\alpha$), then this automorphism is of order 4, and the group is the cyclic group of order 4.

But it could be that every automorphism is its own inverse; if it maps $\alpha$ to some other conjugate, say, to $\eta$, then it maps $\eta$ to $\alpha$, and also interchanges the other two conjugates. Then the group is Klein-4.

In other words, just knowing where $\alpha$ goes doesn't tell you where the other conjugates go, and there are two essentially different possibilities, corresponding to the two groups of order 4.

To get you started on iii), try to find an automorphism $\sigma$ of order 4, and an automorphism $\tau\ne\sigma^2$ of order 2, and try to show they satisfying the defining relation of the dihedral group, $\tau\sigma=\sigma^3\tau$.

EDIT: Maybe an example of ii) will help to see what happens in this case. Take the polynomial $x^4-4x^2+2$. The zeros are $\pm\alpha$ and $\pm\beta$ where $$\alpha=\sqrt{4+2\sqrt2},\quad\beta=\sqrt{4-2\sqrt2}$$ We have $$\alpha^2=4+2\sqrt2,\quad\alpha\beta=2\sqrt2$$ Let $\sigma(\alpha)=\beta$. Then $\sigma(\alpha^2)=\beta^2$, which implies $\sigma(\sqrt2)=-\sqrt2$. Then $$\sigma(\beta)=\sigma(2\sqrt2/\alpha)=-2\sqrt2/\beta=-\alpha$$ and we see that $\sigma$ has order 4.