Let $G$ be a finite nonabelian $p$group such that $G$ contains at least an element of order $p^2$ and for every nontrivial normal subgroup $N$, $G/N$ has not any elements of order p^2 and G/Z(G) is nonabelian group and G/G' is elementary abelian group. 1Is there a group with these properties? 2Are there up to $p$ nontrivial proper subgroups such that intersection of every nontrivial proper subgroup of $G$ with at least one subgroup of these $p$ subgroups is nontrivial?

$\begingroup$ What do you mean by "for every $N$"? Because it cannot include the trivial subgroup. If you mean for every $N\neq 1$, then it follows that $G$ is extraspecial. $\endgroup$ – YCor Oct 28 '16 at 16:51

$\begingroup$ @YCor: Must it? Take the extraspecial group of order $p^3$ and exponent $p$, and adjoin a central $p$th root to its nontrivial central element, $G=\langle x,y,z\mid x^p=y^p=1, [z,x]=[z,y]=1, [x,y]=z^p\rangle$. The center is generated by $z$, and every nontrivial normal subgroup contains $z^p$; the quotient $G/\langle z^p\rangle$ is elementary abelian of rank $3$. But $G$ is not extraspecial. $\endgroup$ – Arturo Magidin Oct 28 '16 at 20:03

$\begingroup$ @ArturoMagidin Yes, sorry I meant a $p$group whose derived subgroup has order $p$ (the center might be larger). Anyway the only other possibility (in this context) is that the center is cyclic of order $p^2$, and it follows that the only nonextraspecial examples are the central product of an extraspecial group with a cyclic group of order $p^2$. $\endgroup$ – YCor Oct 28 '16 at 20:55
This is a strange question but I think the answer is yes, at least for $p>2$. It follows from the second assumption that every nontrivial normal subgroup contains the commutator subgroup, so in particular every order p subgroup of the center ontains the commutator subgroup. Because there always is one, the commutator is order p and central. The quotient by this subgroup is an elementary abelian $p$group.
Let $a$ and $b$ be two elements of order $p$. Then as $a$ and $b$ commute with $[a,b]$, we have $(ab)^p =a^p b^p [a,b]^{p(p1)/2}=[a,b]^{p(p1)/2}$. Because $p>2$ and $[a,b]^p =1$, we have $(ab)^p=1$. So the elements of order $p$ actually form a subgroup. By the first assumption this is proper. Of course every nontrivial subgroup intersects it nontrivially.

$\begingroup$ A similar argument works when $P$ is a $2$group containing an element of order $8$ with the property that $P/N$ is Abelian for each nonidentity normal subgroup $N$ of $P$. For then $P$ is nilpotent of class at most two, and the elements of order dividing $4$ form a subgroup. $\endgroup$ – Geoff Robinson Oct 30 '16 at 13:54