(1) $N(H)$ operates on $H$ by conjugation and precisely $C(H)$ opererates trivially (i.e. is the kernel of the corresponding homomorphism $N(H)\to \operatorname{Aut}(H)$
(2) should be both obvious and actually shown before (1).
EDIT: After (justified) complaints about the use of the word "obvious", let us rewrite to obtain something to answer (1) and (2) at the same time:
Recall that if $\psi\colon A\to B$ is a group homomorphism, then $\ker\psi\lhd A$ and $A/\ker\psi\cong\operatorname{im}\phi.
If $g\in N_G(H)$ and $h\in H$, then by definition $ghg^{-1}\in H$. Since conjugation-with-$g$ is an automorphism of $G$ and leaves $H$ invariant, it is also an automorphism of $H$. Thus we have a group homomorphism $\begin{matrix}\Phi\colon &N_G(H)&\to&\operatorname{Aut}(H)\\&g&\mapsto&(h\mapsto ghg^{-1})\end{matrix}$ What is the kernel of $\Phi$? We have $g\in\ker\Phi$ iff $h=ghg^{-1}$ for all $h\in H$ or equivalently $hg=gh$ for all $h\in H$, i.e. $\ker\Phi = C_G(H)$, thus showing both (1) and (2).