Let $G=S_3$ be the group of permutations on three elements, and $H$ be the subgroup of even permutations. This answers yes to both questions.


Sure, here's an answer to both. Consider the quaternions, $Q_{8}$, and the normal subgroup $H = \langle i \rangle$. Then $Q_{8}/H \cong \mathbb{Z}_{2}$, but $Q_{8}$ is neither cyclic nor abelian.

Here's another more general example. Consider $D_{2n} = \langle r, s \mid r^{n}=s^{2} =1, rs=sr^{-1}\rangle$. The subgroup $H = \langle r \rangle$ is always normal, since $H$ has index $2$ in $G$, and $H$ is always cyclic. Furthermore, $D_{2n}/H \cong \mathbb{Z}_{2}$, and so this again answers both questions, since $D_{2n}$ is not abelian for $n\geq 3$.

You can see this will also generalize to any group with a cyclic subgroup of index $2$.