13. (16 points) Given the hyperbola \( C: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \) \((a>0, b>0)\) with an eccentricity of 2, a line \( l \) passing through the point \( P(0, m) \) \((m>0)\) with a slope of 1 intersects the hyperbola \( C \) at points \( A \) and \( B \), and
\[
\overrightarrow{A P}=3 \overrightarrow{P B}, \overrightarrow{O A} \cdot \overrightarrow{O B}=3 \text {. }
\]
(1) Find the equation of the hyperbola.
(2) Let \( Q \) be a moving point on the right branch of the hyperbola \( C \), and \( F \) be the right focus of the hyperbola \( C \). Does there exist a fixed point \( M \) on the negative half-axis of the \( x \)-axis such that \( \angle Q F M=2 \angle Q M F \)? If it exists, find the coordinates of point \( M \); if not, explain the reason.
13. (1) From the eccentricity of the hyperbola being 2, we know $c=2a, b=\sqrt{3}a$.
Thus, the equation of the hyperbola can be written as $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1$.
Also, the line $l: y=x+m$, when combined with the hyperbola equation, gives
$$
2x^{2}-2mx-m^{2}-3a^{2}=0.
$$

Let points $A(x_{1}, y_{1}), B(x_{2}, y_{2})$. Then
$$
x_{1}+x_{2}=m, x_{1}x_{2}=\frac{-m^{2}-3a^{2}}{2}.
$$

Since $\overrightarrow{AP}=3\overrightarrow{PB}$, we have
$$
(-x_{1}, m-y_{1})=3(x_{2}, y_{2}-m).
$$

Thus, $x_{1}=-3x_{2}$.
Combining with $x_{1}+x_{2}=m$, we solve to get
$$
x_{1}=\frac{3}{2}m, x_{2}=-\frac{1}{2}m.
$$

Substituting into equation (2) gives
$$
\begin{array}{l}
-\frac{3}{4}m^{2}=\frac{-m^{2}-3a^{2}}{2} \Rightarrow m^{2}=6a^{2}. \\
\text{Also, } \overrightarrow{OA} \cdot \overrightarrow{OB}=x_{1}x_{2}+y_{1}y_{2} \\
=x_{1}x_{2}+(x_{1}+m)(x_{2}+m) \\
=2x_{1}x_{2}+m(x_{1}+x_{2})+m^{2} \\
=m^{2}-3a^{2}=3a^{2}=3,
\end{array}
$$

Thus, $a^{2}=1$.
At this point, $m=\sqrt{6}$, substituting into equation (1) and simplifying gives $2x^{2}-2\sqrt{6}x-9=0$.
Clearly, this equation has two distinct real roots.
Therefore, $a^{2}=1$ meets the requirements.
Hence, the equation of the hyperbola $C$ is $x^{2}-\frac{y^{2}}{3}=1$.
(2) Assume point $M(t, 0)$ exists.
From (1), we know the right focus of the hyperbola is $F(2,0)$.
Let $Q(x_{0}, y_{0}) (x_{0} \geqslant 1)$ be a point on the right branch of the hyperbola $C$.
When $x_{0} \neq 2$,
$\tan \angle QFM=-k_{QF}=-\frac{y_{0}}{x_{0}-2}$,
$\tan \angle QMF=k_{QM}=\frac{y_{0}}{x_{0}-t}$.
Since $\angle QFM=2 \angle QMF$, we have
$$
-\frac{y_{0}}{x_{0}-2}=\frac{\frac{2y_{0}}{x_{0}-t}}{1-\left(\frac{y_{0}}{x_{0}-t}\right)^{2}}.
$$

Substituting $y_{0}^{2}=3x_{0}^{2}-3$ into the above equation and simplifying gives
$$
\begin{array}{l}
(4+2t)x_{0}-4t=-2tx_{0}+t^{2}+3 \\
\Rightarrow\left\{\begin{array}{l}
4+2t=-2t, \\
-4t=t^{2}+3
\end{array} \Rightarrow t=-1.\right.
\end{array}
$$

When $x_{0}=2$, $\angle QFM=90^{\circ}$, and when $t=-1$, $\angle QMF=45^{\circ}$, which satisfies $\angle QFM=2 \angle QMF$. Therefore, the point $M(-1,0)$ that meets the conditions exists.