a) Since $G$ is a group, it follows that $|xH| = p$. Therefore
$$n = |G| \ge |H \cup xH| = |H| + |xH| - |H \cap xH| = 2p - |H \cap xH|,$$
whence $|H \cap xH| \ge 2p - n$.

b) Let have $x \in H$ and $y \in H \cap xH$. Then $y = y^{-1}$ and $y = xh$, $h \in H$. Since $xy = x^2h = h \in H$, it follows that $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$. Therefore $x$ commutes with all elements of $H \cap xH$.
Since $|H \cap xH| \ge 2p - n > \frac{3n}{2} - n = \frac{n}{2}$, it follows the subgroup of the elements that commute with $x$ has at least $\frac{n}{2}$ elements. From Lagrange's theorem, it follows that $x$ commutes with all elements of $G$. Therefore $H \subseteq Z(G)$, the center of $G$. It results $|Z(G)| \ge |H| = p > \frac{3n}{4} > \frac{n}{2}$, so $Z(G) = G$.

c) Assume $G$ is commutative.
If $x, y \in H$, then $(xy)^2 = x^2y^2 = e$, so $xy \in H$. Since $H$ is a finite set, it follows that $H$ is a subgroup of $G$.
The condition $|H| > \frac{n}{2}$ forces $H = G$, so $\frac{3n}{4} \ge |H| = |G| = n$, contradiction.