(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 $p$.
11. (20 points) As shown in Figure 4, the ellipse $C_{1}: \frac{x^{2}}{4}+y^{2}=1$, the parabola $C_{2}: x^{2}=2 p y(p>0)$, let $C_{1}$ and $C_{2}$ intersect at points $A$ and $B$, and $O$ be the origin.
(1) From the symmetry of the parabola, ellipse, and circle, we know that the circumcenter of $\triangle A B O$ is the upper vertex $M(0,1)$ of the ellipse. Therefore,
$$
M A=M B=M O=1 \text {. }
$$

Let $B\left(x_{0}, y_{0}\right)\left(x_{0}>0\right)$. Hence,
$$
\left\{\begin{array} { l } 
{ x _ { 0 } ^ { 2 } = 2 p y _ { 0 } , } \\
{ \frac { x _ { 0 } ^ { 2 } } { 4 } + y _ { 0 } ^ { 2 } = 1 , } \\
{ x _ { 0 } ^ { 2 } + ( y _ { 0 } - 1 ) ^ { 2 } = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{0}^{2}=\frac{4(2 \sqrt{13}-5)}{9}, \\
y_{0}=\frac{-1+\sqrt{13}}{3}, \\
p=\frac{7-\sqrt{13}}{6} .
\end{array}\right.\right.
$$
(2) It is easy to see that points $O, A, N, B$ are concyclic.

Let $A B$ intersect the $y$-axis at point $C\left(0, y_{0}\right)$.
By the intersecting chords theorem, we have
$$
\begin{array}{l}
A C \cdot C B=C N \cdot C O \\
\Rightarrow y_{0}\left(\frac{13}{2}-y_{0}\right)=x_{0} x_{0}=2 p y_{0} \\
\Rightarrow y_{0}=\frac{13}{2}-2 p .
\end{array}
$$

Substituting into $x_{0}^{2}=2 p y_{0}$, we get
$$
x_{0}^{2}=2 p\left(\frac{13}{2}-2 p\right) \text {. }
$$

Substituting equations (1) and (2) into the ellipse equation, we get
$$
\begin{array}{l}
\frac{13 p-4 p^{2}}{4}+\left(\frac{13}{2}-2 p\right)^{2}=1 \\
\Rightarrow p=3
\end{array}
$$