Number of subfields of splitting fields of $x^5-5$ over $\mathbb{Q}$.

Since $K=\mathbb{Q}(\zeta,\sqrt[5]{5})$, an element $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ is completely determined by what $\sigma(\zeta)$ and $\sigma(\sqrt[5]{5})$ are. Note that any $\sigma\in\mathrm{Gal}(K/\mathbb{Q})$ must have $$\sigma(\zeta)=\zeta^k\text{ for some }1\leq k\leq 4,\qquad\quad \sigma(\sqrt[5]{5})=\zeta^r\sqrt[5]{5}\text{ for some }0\leq r\leq 4$$ (These comprise the $20$ different elements of $\mathrm{Gal}(K/\mathbb{Q})$.)

To understand $\mathrm{Gal}(K/\mathbb{Q})$, try to write it using a presentation (generators and relations). Can you think of any particular elements of $\mathrm{Gal}(K/\mathbb{Q})$ that, taken together, generate the entire group? Try to treat the two generators of the field "orthogonally" - find one automorphism that permutes the various powers of $\zeta$ and ignores $\sqrt[5]{5}$, and another automorphism that permutes the various 5th roots of $5$ and ignores any $\zeta$s (or at least, any $\zeta$s that aren't part of one of the 5th roots of $5$). Then combinations of these two should allow you to produce any desired effect on both $\zeta$ and $\sqrt[5]{5}$.

Then, determine the relations between those generating elements. That should make it easier to figure out the subgroups of $\mathrm{Gal}(K/\mathbb{Q})$ (and hence, the subfields of $K$).

This group of order $20$ is isomorphic to the following group of $2\times2$ matrices over the field $k=\Bbb{F}_5$ $$ G=\operatorname{Gal}(K/\Bbb{Q})\simeq\left\{ \left(\begin{array}{cc}a&b\\0&1\end{array}\right)\big\vert\,a,b\in k, a\neq0 \right\}. $$ This isomorphism comes from mapping the automorphism $\sigma:\zeta_5\mapsto\zeta_5, \root5\of5\mapsto\zeta_5\root5\of5$ to the matrix $\pmatrix{1&1\cr0&1\cr}$, and the automorphisms $\tau_a:\root5\of5\mapsto\root5\of5, \zeta_5\mapsto \zeta_5^a$, $a=1,2,3,4$, to the diagonal matrices of $G$. Do check that this extends to an isomorphism. Alternatively you can bijectively map the zeros $x^5-5$ to elements of $k$, and then check that the automorphisms of $K$ are the linear bijections $x\mapsto ax+b$ from $k$ to itself.

You can then list the subgroups systematically by their orders. The Sylow $5$-subgroup consists of powers of $\sigma$, and it is a normal, hence unique subgroup of its order. The diagonal matrices of $G$ form its (cyclic) Sylow $2$-subgroup. It is equal to its own normalizer and thuse has five conjugates. This may be easier to see from the above realization of $G$ as permutations of $k$ - the Sylow $2$-subgroups are then exactly the point stabilizers.

This should give you enough structure to enable you to also list the subgroups of orders two and ten.