Let $F$ be any field. The compositum of separable extensions of $F$ contained in the algebraic closure $\overline{F}$ of $F$ will itself be separable, and so there is a largest separable extension of $F$ contained in $\overline{F}$ (namely, the compositum of all separable extensions). This is called the separable closure of $F$ in $\overline{F}$. (See for example Lang's Algebra, revised 3rd Edition, Theorem 4.5 and discussion following, pp. 241f).
Now start with a non-perfect field; for example, take $\mathbb{F}_p(x)$, the field of rational functions with coefficients in the field of $p$ elements.
Let $K$ be the separable closure of $F$ as above; because $F$ is not perfect, $K$ cannot equal $\overline{F}$. In particular, $K$ is not algebraically closed.
However, every nontrivial algebraic extension of $K$ is not separable (in fact, it will be purely inseparable): because if $L$ is an algebraic separable extension of $K$, then $L$ is also an algebraic separable extension of $F$, hence must be contained in $K$, so $L=K$.
Thus, no nontrivial algebraic extension of $K$ is separable, so no nontrivial algebraic extension of $K$ is Galois over $K$; and yet there are nontrivial algebraic extensions of $K$, since $K$ is not algebraically closed.
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 $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
Best Answer
There are infinite algebraic Galois extensions of $\mathbb{Q}$, simply take a splitting field $F$ of a infinite family of polinomials like $x^2-p$ where $p\in\mathbb{Z}$ is a prime. Remember that, since $\mathbb{Q}$ has characteristic zero every extension is separable, and a splitting field of a family of polynomials is normal, so is Galois.
Now, if $K$ is a splitting field of a (only one) polynomial $p(x)\in\mathbb{Q}[x]$, then $K/\mathbb{Q}$ is finite. In fact, using basic Galois Theory $[K:\mathbb{Q}]\leq n!$, where $n=\deg p(x)$.
Edit: In the last question. Using Galois theory, there are finitely many intermediate fields as there are finite subgroups.