= Solution
For the first field, set $K=\mathbb Q(\sqrt5,i)$. Its <field extension degree> is four, since $\mathbb Q(\sqrt5)$ is real and adjoining $i$ doubles its degree. The <polynomial> $X^3-5$ is irreducible over $K$: a reducible cubic would have a root in $K$, but every root has degree three over $\mathbb Q$ by <Eisenstein criterion>, impossible by the <tower law> because $3\nmid4$. Thus
$$
\boxed{[\mathbb Q(\sqrt[3]5,\sqrt5,i):\mathbb Q]=12}.
$$
For the second field, put $\alpha=\sqrt[4]5>0$. The <splitting field> is $F=\mathbb Q(\alpha,i)$ and has degree eight, since $X^4-5$ is Eisenstein and $\mathbb Q(\alpha)$ is real. Define <automorphisms> $r(\alpha)=i\alpha,r(i)=i$ and $s(\alpha)=\alpha,s(i)=-i$. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$, and give eight distinct automorphisms. Hence \b[$\operatorname{Gal}(F/\mathbb Q)\cong D_8$.]
The <Galois correspondence> gives all the subfields in the following table. The middle column records generators, not merely their degrees.
$$
\begin{array}{c|c|c}
\text{subgroup fixing the field}&\text{field}&\text{degree over }\mathbb Q\\ \hline
\langle r,s\rangle&\mathbb Q&1\\
\langle r\rangle&\mathbb Q(i)&2\\
\langle r^2,s\rangle&\mathbb Q(\sqrt5)&2\\
\langle r^2,rs\rangle&\mathbb Q(i\sqrt5)&2\\
\langle r^2\rangle&\mathbb Q(\sqrt5,i)&4\\
\langle s\rangle&\mathbb Q(\alpha)&4\\
\langle r^2s\rangle&\mathbb Q(i\alpha)&4\\
\langle rs\rangle&\mathbb Q((1+i)\alpha)&4\\
\langle r^3s\rangle&\mathbb Q((1-i)\alpha)&4\\
\{1\}&F&8
\end{array}
$$
Each displayed generator is fixed by the indicated subgroup. The two last quartic generators have fourth power $-20$, so their degree is four by <Eisenstein criterion>; the other quartic degrees are immediate. These are all the subgroups of $D_8$ and hence all the subextensions.
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