Let's use the following numbering on the roots of your polynomial: $$\{z_1=\root3\of5,z_2=ωz_1,z_3=ω^2z_1,z_4=\sqrt3,z_5=−\sqrt3\}.$$ Any automorphism $\sigma$ of the splitting field must permute these numbers as they are the zeros of your polynomial. As the splitting field $F$ is generated by them, the automorphism $\sigma$ is fully determined once we know $\sigma(z_j),j=1,2,3,4,5.$ This gives us the usual way of identifying $\sigma$ with an element of $S_5=\operatorname{Sym}(\{z_1,z_2,z_3,z_4,z_5\})$.
But there are further constraints. The numbers $z_1,z_2,z_3$ are zeros of the factor $x^3-5$. As that factor has rational coefficients, all automorphisms must permute these three roots among themselves. Similarly for the remaining pair $z_4,z_5$ as they are the zeros of $x^2-3$. Therefore in the identification of an automorphism with a permutation in $S_5$ of the previous paragraph only the permutations in $\textrm{Sym}(\{z_1,z_2,z_3\})\times \textrm{Sym}(\{z_4,z_5\})$ are allowed. There are 12 such permutations forming a group isomorphic to $S_3\times S_2$. As you had established by other means that $[F:\mathbb{Q}]=12$, we can conclude that all such permutations come from actual automorphisms, and thus the Galois group is isomorphic to $G=S_3\times S_2$.
Let us first calculate the fixed field of the subgroup $H_1=\langle\sigma\rangle$, where $\sigma=(123)(45)$. Here we are given that $\sigma(z_1)=z_2$ and that $\sigma(z_2)=z_3$. As $\sigma$ is an automorphism of fields we get
$$
\sigma(\omega)=\sigma\left(\frac{z_2}{z_1}\right)=\frac{\sigma(z_2)}{\sigma(z_1)}=\frac{\omega^2z_1}{\omega z_1}=\omega.
$$
Therefore $\omega$ is fixed by $\sigma$, and therefore also by any power of $\sigma$.
Thus $\mathbb{Q}(\omega)\subseteq Inv(H_1)$. From Galois theory we know that $[Inv(H_1):\mathbb{Q}]=|G|/|H|=12/6=2$. As $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ it follows that $\mathbb{Q}(\omega)=Inv(H_1)$.
It is fairly clear that the fixed field of the subgroup $H_2\simeq S_3$ that keeps both $z_4$ and $z_5$ fixed is $Inv(H_2)=\mathbb{Q}(\sqrt3)$.
Let's try the last subgroup $H_3$ of order 6.
I describe it as follows. We have a homomorphism $f$ from $S_3=Sym(\{z_1,z_2,z_3\})$ to
$S_2=\textrm{Sym}(\{z_4,z_5\})$ that maps a permutation $\alpha\in S_3$ to the identity element $(4)(5)$ (resp. to the 2-cycle $(45)$) according to whether $\alpha$ is an even (resp. odd) permutation. Then
$$
H_3=\{(\alpha,f(\alpha))\in S_3\times S_2\mid \alpha\in S_3\}.
$$
We easily see that $H_3$ is generated by $\beta=(123)$ and $\gamma=(12)(45)$.
Because $\beta$ acts on the set $\{z_1,z_2,z_3\}$ the same way as $\sigma$ above, we see that $\beta(\omega)=\omega$ and also that $$\beta(\sqrt{-3})=\beta(2\omega+1)=2\omega+1=\sqrt{-3}.$$ Because $\beta(z_4)=z_4$, we obviously also have $\beta(\sqrt3)=\sqrt3$. What about $\gamma$? First we get
$$
\gamma(\omega)=\gamma(\frac{z_2}{z_1})=\frac{\gamma(z_2)}{\gamma(z_1)}=\frac{z_1}{z_2}=\omega^2.
$$
As $2\omega=-1+\sqrt{-3}$ and $2\omega^2=-1-\sqrt{-3}$, this implies that $\gamma(\sqrt{-3})=-\sqrt{-3}$.
But we also have
$$
\gamma(\sqrt3)=\gamma(z_4)=z_5=-\sqrt3.
$$
Putting all these bits together we see that
$$
\gamma(i)=\gamma\left(\frac{\sqrt{-3}}{\sqrt3}\right)=\frac{-\sqrt{-3}}{-\sqrt3}=i.
$$
Similarly we see that $\beta(i)=i$. We can then conclude that $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
Let $E=\mathbb{Q}(\sqrt[7]{5})$ and suppose that you have another field $F$ of degree 7. Then $[F(\sqrt[7]{5}):F][F:\mathbb{Q}]\leq 14$ so we must have $[F(\sqrt[7]{5}):F]=2$. It follows that $\alpha=\sqrt[7]{5}$ satisfies an irreducible polynomial of degree 2 over $F$.
There is now a general fact that states that if a polynomial $x^p-a$ is reducible over some field F for some $p$ prime, then it has a root in F.
More precisely, this polynomial is of the form $(x-\zeta^i \alpha)(x-\zeta^j \alpha)=x^2 -(\zeta^i+\zeta^j)\alpha +\zeta^{i+j}\alpha^2$ where $\zeta$ is a primitive 7 root of unity and $0\leq i<j<7$ - this is because it divides $x^7-5=\prod_{i=1}^7 (x-\zeta^i \alpha)$.
You now have that $\zeta^{i+j}\alpha^2 \in F$ so by taking the 4th power you also have that $\zeta^{4i+4j}\alpha^8=5\zeta^{4i+4j}\alpha \in F$ and therefore $\zeta^{4i+4j}\alpha \in F$. We now have two options - either $\zeta^{4i+4j}=1$ but then $E=F$, or it is some primitive 7 root of unity, which then must be in $F(\alpha)$. But this means that $6\mid [F(\alpha):\mathbb{Q}]$ - contradiction.
Best Answer
You are correct in that it is not a splitting field of $\sqrt[4]{2}$, since in particular it is entirely real, but $X^4-2$ is irreducible with complex solutions. Thus it is not a splitting field of any polynomial, since if it were then the extension would be Galois and contain all the roots of $X^4-2$.
One perfectly legitimate strategy would be to look the subfields of some Galois extension $L/\mathbb{Q}$ containing $\mathbb{Q}(\sqrt[4]{2})$ (which can be easily determined once you identify the Galois group) and to identify those that lie inside $\mathbb{Q}(\sqrt[4]{2})$, by looking at the assosciated subgroups of $Gal(L/\mathbb{Q})$ containing $Gal(L/\mathbb{Q}(\sqrt[4]{2}))$. However, this is a lot of work (and requires a lot of theory, although this is not a bad thing in and of itself). There is a much simpler and more elementary way to approach this problem, though.
Hint:
$[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] = 4$ since $f(X) = X^4-2$ is irreducible, so the only proper subfields are degree $2$ over $\mathbb{Q}$. If $K$ is one such, let $g(X)$ be the minimal polynomial of $\sqrt[4]{2}$ over $K$. $g$ must have degree $2$, divide $f$ and have $\sqrt[4]{2}$ as a root. This gives a small number of possible options. We can then use the fact that coefficients of $g$ lie in $K$ (which is a real field) to determine what possible values $g$ may take, and this tells us enough to be able to determine all possible subfields $K$.
It is worth noting that the actual solution is shorter than the hint, but I think it's a good exercise to work through if you haven't seen it before.