Jyrki has already typed up a great answer showing how to compute the fixed fields. In my post I will concentrate on how to compute the Galois group. Let us write $E = \Bbb{Q}(\sqrt{3},i,\sqrt[3]{5}) = \Bbb{Q}(\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_4)$ and $F = \Bbb{Q}$ where
$$\begin{eqnarray*} \alpha_1 &=& \sqrt[3]{5} \\
\alpha_2 &=& \omega\sqrt[3]{5} \\
\alpha_3 &=& \omega^2 \sqrt[3]{5} \\
\alpha_4 &=& \sqrt{3} \\
\alpha_5 &=& -\sqrt{3}\\
\end{eqnarray*}$$
and $\omega = e^{e\pi i/3}$. Now what you said about "complex conjugation" and concluding from there that the Galois group was $D_{12} \cong S_3 \times S_2$ and not $A_4$ is not right. I think you meant by complex conjugation the automorphism that exchanges the roots of $x^2- 2$? In any case what I will do now is use my bare hands to compute the Galois group. Then you will be $10^6$ percent sure that it must be $D_{12}$ of order 12. I will now bash out the possibilities for the action of a $\sigma \in \operatorname{Gal}(E/F)$ on the roots (in other words, bash out all possibilities for automorphisms). We will write for these automorphisms cycles of $S_5$ since we can view $\operatorname{Gal}(E/F)$ as a subgroup of $S_5$. For example, the cycle $(123)$ sends $\alpha_1 \mapsto \alpha_2, \alpha_2 \mapsto \alpha_3$ and $\alpha_3 \mapsto \alpha_1$. $\alpha_4$ and $\alpha_5$ are fixed. The elements are:
$$ e = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_2 \\ \alpha_3 \mapsto \alpha_3 \\ \alpha_4 \mapsto \alpha_4 \\ \alpha_5 \mapsto \alpha_5 \end{cases} \hspace{4mm} (45) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_2 \\ \alpha_3 \mapsto \alpha_3 \\ \alpha_4 \mapsto \alpha_5 \\ \alpha_5 \mapsto \alpha_4 \end{cases} \hspace{4mm} (23)(45) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_3 \\ \alpha_3 \mapsto \alpha_2 \\ \alpha_4 \mapsto \alpha_5 \\ \alpha_5 \mapsto \alpha_4 \end{cases} \hspace{4mm} (23) = \begin{cases} \alpha_1 \mapsto \alpha_1 \\ \alpha_2 \mapsto \alpha_3 \\ \alpha_3 \mapsto \alpha_2 \\ \alpha_4 \mapsto \alpha_4 \\ \alpha_5 \mapsto \alpha_5 \end{cases}$$
and the rest of the cycles are $(123),(123)(45),(12),(12)(45),(132)(45),(132),(13),(13)(45)$. Now I explain a bit more on how these were calculated. You start say with $\alpha_1$. Then if it's sent to itself, then you have only two choices for where $\alpha_2$ is sent to. Either itself, or $\alpha_3$. We don't need to do this $\alpha_3$ as where $\alpha_2$ is sent to already determines this. Then you deal with $\alpha_4$ and $\alpha_5$. Keep going and eventually get all the elements I talked about above. Each of those is a valid automorphism of $E/F$ and since the Galois group has 12 elements, this must be all of them.
If we set $x = (123)(45)$ and $y = (12)$ we see that we have the elements of $\operatorname{Gal}(E/F)$ being
$$\{e,x,x^2 ,x^3,x^4, x^5, y,xy, x^2y ,x^3y,x^4, x^5y\}$$
that satisfy the relations $x^6 = y^2 = 1$ and as you can check we have $yx = x^5y = x^{-1}y$. Hence we conclude immediately from here that $\operatorname{Gal}(E/F)$ must be isomorphic to $D_{12}$ because the structure of the elements and the relations that they satisfy are exactly those of $D_{12}$.
Now I see you have some trouble finding the subgroup that fixes $\Bbb{Q}(i)$. Here is how you can do it. Your three subgroups of order 6 are:
$$\langle x \rangle, \{e,x^2,x^4,y,x^2y,x^4, \}, \{e,x^2,x^4, xy,x^3y,x^5y\}.$$
Now I don't think the first subgroup will fix $i$ because it kinda "moves everything around. My guess is that the last one in the right is the one that fixes $i$. We check this. Recall that
$$x^2 = (132),x^4= (123), xy = (13)(45),x^3y = (12)(45),x^5y = (23)(45)$$
and from the fact that $\alpha_2 = \omega \sqrt[3]{5} = \left(\frac{-1 + i\sqrt{3}}{2}\right)\sqrt[3]{5}$ we get that $i = \frac{1}{\alpha_4}\left(\frac{2\alpha_2}{\alpha_1}+ 1\right)$. Now if you apply each element of the subgroup to this and noting that $\omega^2 = \omega^{-1}$ you will easily see that this fixes $i$. For example, if you let $xy = (13)(45)$ act on $i$ we have that
$$\begin{eqnarray*} (13)(45) \Bigg( \frac{1}{\alpha_4}\left( \frac{2\alpha_2}{\alpha_1} +1 \right) \Bigg) &=& \frac{1}{\alpha_5}\left( \frac{2\alpha_2}{\alpha_3} +1 \right) \\
&=& \frac{1}{\alpha_5}(2\omega^2 + 1)\\
&=& \frac{1}{\alpha_5}(-i\sqrt{3})\\
&=& i.
\end{eqnarray*}$$
You should be able to bash out the rest. If you do this the smart way like what Jyrki has done you just need to show that $i$ is fixed by just the generators of this subgroup, namely $(123)$ and $(12)(45)$. Does this help you?