13. Given the parabola $C: y=a x^{2}(a>0)$, the line $y=x+2$ intersects the parabola at points $A$ and $B$, and $M$ is the midpoint of segment $AB$. 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 $AB$.
(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 the reason.
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 $x_{1}+x_{2}=\frac{1}{a}, x_{1} x_{2}=-\frac{2}{a}$.
Thus, $x_{N}=x_{M}=\frac{x_{1}+x_{2}}{2}=\frac{1}{2 a}, y_{N}=a x_{N}^{2}=\frac{1}{4 a}$.
From $y^{\prime}=\left(a x^{2}\right)^{\prime}=2 a x$, we know that the slope of the tangent line $l$ to the parabola $C$ at point $N$ is $k_{l}=2 a \cdot \frac{1}{2 a}=1$.

Therefore, the tangent line $l$ to the parabola $C$ at point $N$ is parallel to the line $A B$.
(2) Suppose there exists a real number $a$ such that $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$. Since $M$ is the midpoint of segment $A B$, we have
$$
\overrightarrow{N A} \cdot \overrightarrow{N B}=0 \Leftrightarrow N A \perp N B \Leftrightarrow|M N|=\frac{1}{2}|A B| \text {. }
$$

Since $M N \perp x$-axis, we have
$$
\begin{array}{l}
|M N|=\left|\frac{1}{2 a}+2-\frac{1}{4 a}\right|=\frac{1}{4 a}+2 . \\
\text { Also, }|A B|=\sqrt{2} \cdot\left|x_{1}-x_{2}\right| \\
=\sqrt{2} \cdot \sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}=\sqrt{2} \cdot \sqrt{\frac{1}{a^{2}}+\frac{8}{a}},
\end{array}
$$

Then $\left(\frac{1}{4 a}+2\right)^{2}=\frac{1}{4} \times 2\left(\frac{1}{a^{2}}+\frac{8}{a}\right)$.
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 $\overrightarrow{N A} \cdot \overrightarrow{N B}=0$.