(13 Given the parabola $C: y=a x^{2}(a>0)$, the line $y=x+2$ intersects the parabola $C$ at points $A$ and $B$, and $M$ is the midpoint of segment $A B$. A vertical line through $M$ intersects the parabola $C$ at point $N$.
(1) Prove: The tangent line $l$ to the parabola $C$ at point $N$ is parallel to $A B$.
(2) Does there exist a real number $a$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$? If it exists, find the value of $a$; if not, explain why.
(13) (1) From $\left\{\begin{array}{l}y=x+2, \\ y=a x^{2}\end{array}\right.$ we get $a x^{2}-x-2=0$. Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then
$$
x_{1}+x_{2}=\frac{1}{a}, x_{1} x_{2}=-\frac{2}{a} .
$$

Therefore,
$$
\begin{array}{l}
x_{N}=x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{1}{2 a}, \\
y_{N}=a \cdot x_{N}^{2}=\frac{1}{4 a} .
\end{array}
$$

From $y^{\prime}=\left(a x^{2}\right)^{\prime}=2 a x$, we know that the slope $k_{1}$ of the tangent line $l$ to the parabola $C$ at point $N$ is $2 a$. $\frac{1}{2 a}=1$. Therefore, the tangent line $l$ to the parabola $C$ at point $N$ is parallel to the line $A B$.
(2) Assume there exists a real number $a$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$. Since $M$ is the midpoint of segment $A B$,
$$
\overrightarrow{N A} \cdot \overrightarrow{N B}=0 \Leftrightarrow N A \perp N B \Leftrightarrow|M N|=\frac{1}{2}|A B| .
$$

Since $M N \perp x$-axis,
$$
|M N|=\left|\frac{1}{2 a}+2-\frac{1}{4 a}\right|=\frac{1}{4 a}+2 .
$$

Also,
$$
\begin{aligned}
|A B| & =\sqrt{2} \cdot\left|x_{1}-x_{2}\right| \\
& =\sqrt{2} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
& =\sqrt{2} \cdot \sqrt{\frac{1}{a^{2}}+\frac{8}{a}} .
\end{aligned}
$$

Therefore,
$$
\left(\frac{1}{4 a}+2\right)^{2}=\frac{1}{4} \times 2 \times\left(\frac{1}{a^{2}}+\frac{8}{a}\right),
$$

Solving this, we get $a=\frac{7}{8}$ or $a=-\frac{1}{8}$ (discard). Therefore, there exists a real number $a=\frac{7}{8}$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$.