Galois Groups and intermediate field extensions

For justifying the existence of a unique Sylow-$3$ subgroup, note that such a subgroup is unique if and only if it is normal, and a normal Sylow-$3$ subgroup would correspond to a Galois subextension $ K/\mathbf Q $ with degree $ 8 $. You already know a degree $ 8 $ subextension of $ E/\mathbf Q $, namely $ \mathbf Q(\sqrt[4]{3}, i)/\mathbf Q $, so all you need to show is that this subextension is Galois. This trivially follows from it being the splitting field of $ X^4 - 3 $.

For the last problem, you know that $ [E_f : \mathbf Q] = 6 $ and $ [E_g : \mathbf Q] = 8 $, and yet their compositum has degree $ 24 $. This means $ [E_f \cap E_g : \mathbf Q] = 2 $, and in fact armed with this result you can determine $ E_f \cap E_g = \mathbf Q(\sqrt{-3}) $. The subgroup $ K \subset G $ corresponding to this subfield has all of the required properties.

Since the extensions $ E_f $ and $ E_g $ become linearly disjoint over $ E_f \cap E_g $, you can in fact see from this result that there is a short exact sequence

$$ 0 \to C_3 \times C_2 \times C_2 \to G \to C_2 \to 0 $$