= Solution
{c}
For $h\in H$, define $\pi(h)(n)=hnh^{-1}$. Since $N$ is a <normal subgroup>, this sends $N$ into itself. It preserves multiplication, and its inverse is conjugation by $h^{-1}$, so it is an <group automorphism> of $N$. Moreover
$$
\pi(h_1h_2)(n)=h_1h_2n h_2^{-1}h_1^{-1}
=\bigl(\pi(h_1)\circ\pi(h_2)\bigr)(n).
$$
Thus $\pi:H\to\operatorname{Aut}(N)$ is a <group homomorphism>, induced by the <group action> of <conjugation>.
If $\pi$ is trivial, every element of $H$ commutes with every element of $N$. Hence
$$
F:N\times H\longrightarrow G,\qquad F(n,h)=nh
$$
is a <group homomorphism>, since $(n_1h_1)(n_2h_2)=n_1n_2h_1h_2$. Its <image of a group homomorphism> is a <subgroup> containing both $N$ and $H$, and every product $nh$ lies in $\langle N,H\rangle$, so its image is exactly $\langle N,H\rangle$. If $nh=e$, then $n=h^{-1}\in N\cap H=\{e\}$, so $n=h=e$. Thus $F$ is injective and gives
$$
\boxed{\langle N,H\rangle\cong N\times H.}
$$
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