What are the subfields of $\mathbb Q(e^{\frac{i\pi}{4}})$ ?
Since $\displaystyle e^{\frac{i\pi}{4}}=\frac{\sqrt2+\sqrt{-2}}2$
So $\mathbb Q(\sqrt2), \mathbb Q(\sqrt{-2})$
and $\sqrt2\cdot\sqrt{-2}=\sqrt{-4}=2i$
Therefore $\mathbb Q(i)$ is another one
The problem is also, it is not written that the subfields must contain $\mathbb Q$
Are there finitely many then ?
Here is a solution which is elementary and avoids the use of Galois theory.
Let $K = \mathbb{Q}(e^{i\pi/4})$. Give an argument that $K = \mathbb{Q}(i,\sqrt{2})$. Show that $[K:\mathbb{Q}]=4$.
If $F$ is a subfield of $K$, then $F$ must contain $\mathbb{Q}$. Furthermore, $d=[F:\mathbb{Q}]$ must divide $[K:\mathbb{Q}] = 4$. Thus, $d=1,2,\text{ or }4$. If $d=4$ then $F=K$ and if $d=1$ then $F=\mathbb{Q}$.
It remains to understand the case when $d=2$. In this case, since $F$ is a quadratic field over $\mathbb{Q}$ we can say that $F = \mathbb{Q}(\sqrt{n})$ where $n$ is a square-free integer. Now use this to find the possible values of $n$.