Let $\mathbb{Q}(i)$ be given, where $i^2=-1$. I want to know:
Whether there is a Galois extension $K/\mathbb{Q}(i)$ such that $\mathrm{Gal}(K/\mathbb{Q}(i))\cong D_8$, where $D_8$ is the dihedral group of size 8?
I know the Fundamental Theorem of Galois theory, and how to calculate Galois groups.
I tried several guesses, such as $\mathbb{Q}(2^{1/4},i)$, $\mathbb{Q}(2^{1/8},i)$, but none worked.
You only need to know whether one exists, not what it is, so no need to guess at what an actual extension would be.
Instead, you could think only about the Galois groups. You can use the Fundamental Theorem of Galois Theory to show that such an extension exists if and only if there exists a group $G$ such that $G\cong D_8$ and $\operatorname{Gal}(K/\mathbb{Q}(i))$ is a normal subgroup of $G$. If you show that, and find such a group (up to isomorphism), then you're done.
Why a normal subgroup in particular? Well, it has to be a Galois extension, not just any extension.