Let $G$ be a finite group. Denote by $N = O^{2'}(G)$ the smallest normal subgroup, such that $G / N$ has odd order. Let $H$ a subgroup of odd order which is normalised by every Sylow $2$-subgroup of $G$, then $H$ is normalised by $N$ and further $H \cap N \unlhd N$, so that $H \cap N \le O_{2'}(N)$, where $O_{2'}(N)$ denotes the largest normal subgroup of $N$ of odd order.
Also if $O_{2'}(G) = 1$, then $O_{2'}(N) = 1$.
My questions:
1) Why does $H \cap N \le O_{2'}(N)$ holds?
2) How to proof that if $H$ is normalised by every Sylow-$2$-subgroup, then it is normalised by $N$?
3) Why does $O_{2'}(G) = 1$ implies $O_{2'}(N) = 1$?