11. As shown in the figure, the ellipse $C_{1}: \frac{x^{2}}{4}+y^{2}=1$, the parabola $C_{2}: x^{2}=2 p y(p>0)$, let $C_{1}, C_{2}$ intersect at points $A, B$, and $O$ be the origin.
(1) If the circumcenter of $\triangle A B O$ lies on the ellipse, find the value of the real number $p$;
(2) If the circumcircle of $\triangle A B O$ passes through the point $N$
$\left(0, \frac{13}{2}\right)$, find the value of the real number $p$.
(1) As shown in the figure, by symmetry, the circumcenter of $\triangle A B O$ is
$C(0,1)$, solving the system $\left\{\begin{array}{l}x^{2}+(y-1)^{2}=1 \\ x^{2}+4 y^{2}=4\end{array}\right.$,
$$
\Rightarrow 3 y^{2}+2 y-4=0 \Rightarrow y=\frac{\sqrt{13}-1}{3} \text {. }
$$

Thus, $x^{2}=\frac{8 \sqrt{13}-20}{9}$, substituting into $x^{2}=2 p y$ gives
$$
\frac{8 \sqrt{13}-20}{9}=2 p \cdot \frac{\sqrt{13}-1}{3} \Rightarrow p=\frac{4 \sqrt{13}-10}{3(\sqrt{13}-1)}=\frac{7-\sqrt{13}}{6} \text {. }
$$
(2) From (1), the equation of the circumcircle of $\triangle A B O$ is $x^{2}+y^{2}-\frac{13}{2} y=0$, solving the system $\left\{\begin{array}{l}x^{2}+y^{2}-\frac{13}{2} y=0, \\ x^{2}+4 y^{2}=4\end{array} \Rightarrow 3 y^{2}+\frac{13}{2} y-4=0 \Rightarrow y=\frac{1}{2}\right.$ (negative value discarded).

Thus, $x^{2}=3$, substituting into $x^{2}=2 p y$ gives $p=3$.