Is this homomorphism surjective? For part b) I understand how to show there is a homomorphism from D2p to H via the method described, but why can we say this homomorphism is surjective?


 A: Let $H$ be a group such that there exists a surjective homomorphism $f: D_{2p}\to H$. By the first isomorphism theorem, $$H\cong D_{2p}/\ker f.$$
Thus any such group will have the form $D_{2p}/N$ where $N$ is a normal subgroup of $D_{2p}$.
Conversely, if you have a group of the form $D_{2p}/N$ ($N$ a normal subgroup of $D_{2p}$, you clearly have a surjective homomorphism $$D_{2p}\to D_{2p}/N$$ (the projection).
A: For a homomorphism $\phi : G \rightarrow H$ to be surjective, we need $\mathrm{im}\, \phi = H$. By the first isomorphism theorem, we have $$\frac{G}{\mathrm{ker}\, \phi} \cong \mathrm{im}\, \phi = H.$$ So all such possible $H$ (up to isomorphism) are determined by our 'options' for $\mathrm{ker}\, \phi$. It turns out that every single normal subgroup $N$ of $G$ can arise as the kernel of some homomorphism (can you see which one?), and the kernel of every homomorphism of $G$ is of course a normal subgroup of $G$ - so these two things are in complete correspondence. 
Therefore, all such $H$ are given by $G/N$ as $N$ runs over all isomorphism classes of normal subgroups of $G$.
