Isn't the answer to 1 clearly yes, because the normalizer of the inverse image of a subgroup of a quotient properly contains it?
As for 2, let's fix a prime $p$, and let $P_n$ be a sequence of finite $p$-groups of unbounded derived length. For example, we could take iterated wreath products. Now let $P$ be the direct product of the $P_n$. Then $P$ is not solvable, since there is no bound on its derived length.
To show that $P$ satisfies the normalizer condition, let $Q$ be a subgroup of $P$. If $Q$ does not contain $Z(P)$, which is the direct product of the $Z(P_n)$, then $H$ is properly contained in $QZ(P) \le N_P(Q)$. So $Z(P) \le Q$. But now if $P$ does not contain the second centre $Z_2(P)$ of $P$, then $Q$ is properly contained in $QZ_2(P) \le N_P(Q)$. So by induction we get $Z_r(P) \le Q$ for all $r$, but the union of the $Z_r(P)$ is $P$, so $Q=P$.