If $\chi\in\operatorname{Irr}(G)$, $N\unlhd G$, and $\langle\chi_{N},1_{N}\rangle\ne 0$, then $N\subset \operatorname{Ker}(\chi)$. 
Let $N \unlhd G$ and $\chi \in \operatorname{Irr}(G)$. Suppose that $\langle\chi_{N},1_{N}\rangle\ne 0$. Show that $N\subset \operatorname{Ker}(\chi)$.
Hint: Use that, for any character $\theta$ of $H \leq G$, we have $$\operatorname{Ker}(\theta^{G})=\bigcap_{x \in G}\left(\operatorname{Ker}(\theta)\right)^{x}$$

I have no idea about that!

According to Frobenius' reciprocity $[\chi_N,1_N]= [\chi,1_N^G]$, whence $[\chi,1_N^G] \neq 0$, that is $\chi$ is an irreducible constituent of $1_N^G$. This implies that $ker(1_N^G) = core_G(N) =$ (since $N$ is normal) $N \subset ker(\chi)$. (Use that the kernel of a character is the intersection of the kernels of its irreducible constituents.) And $ker(\theta^G)=core_G(ker(\theta))$ is just another way to notate your intersection, it is the largest normal subgroup of $G$ contained in $ker(\theta)$.
