= Solution
The <first isomorphism theorem for groups> states that for a <group homomorphism> $f:G\to Q$,
$$
G/\ker f\cong\operatorname{im}f.
$$
The map $g\ker f\mapsto f(g)$ is well defined because two representatives differ by an element of the kernel; it is a surjective homomorphism onto the image and is injective because $f(g)=e$ exactly when $g\in\ker f$.
For $n_i\in N$ and $h_i\in H$, normality of $N$ gives
$$
(n_1h_1)(n_2h_2)=n_1(h_1n_2h_1^{-1})h_1h_2\in NH,
$$
and $(nh)^{-1}=h^{-1}n^{-1}h\,h^{-1}\in NH$. Hence $NH$ is a subgroup. Also $N\cap H\trianglelefteq H$, since conjugation by $H$ preserves both $N$ and $H$, and $N\trianglelefteq NH$ because $NH\leq G$.
Apply the first isomorphism theorem to $f:H\to NH/N$, $f(h)=hN$. Its kernel is $N\cap H$, and it is surjective because $nhN=hN$. Therefore
$$
\boxed{H/(N\cap H)\cong NH/N}.
$$
If $K,H\trianglelefteq G$, then $KH$ is a normal subgroup: it is a subgroup by the calculation above, and $g(KH)g^{-1}=(gKg^{-1})(gHg^{-1})=KH$. Without normality a product need not be a subgroup. In $S_3$, take $H=\langle(12)\rangle$ and $K=\langle(23)\rangle$. The set $KH$ has four elements, so it cannot be a subgroup of the six-element group by <Lagrange's theorem for finite groups>.
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