1) The fact that $H$ and $N$ have relatively prime orders means their intersection is trivial by Lagrange's theorem. It suffices to prove that $|HN|=|G|$. Suppose not. Then there are $h_0,h_1,n_0,n_1$ with $h_0n_0=h_1n_1$. This means $h_1^{-1}h_0=n_1n_0^{-1}$, and because $H\cap N=\{e\}$, we have $h_0=h_1$ and $n_0=n_1$, which gives the needed contradiction.
2) Suppose $g(N)=K\neq N$. Consider the set $KN$, which is a subgroup as $N$ is normal. The second isomorphism theorem gives $KN/N \cong K/(K\cap N)$ and $|KN||K\cap N| = |N||K|=|N|^2$ and $|K\cap N|$ divides $|N|$ as it is a subgroup, so $|N|$ divides $|KN|$, and we know $|KN|$ divides $|G|$, so $|KN|/|N|$ divides $|G/N|$. But note that $|KN|/|N|$ divides $|N|$ from the prior equation, so we have a contradiction unless $KN=N$ and $K=N$.