14. 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) If \( Q \) is a moving point on the right branch of the hyperbola \( C \), and \( F \) is the right focus of the hyperbola \( C \), does there exist a fixed point \( M \) on the negative half 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.
14. (1) From the eccentricity of the hyperbola being 2, we know that $c=2a, b=\sqrt{3}a$.
Thus, the equation of the hyperbola becomes $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1$.
The equation of the line $l$ is $y=x+m$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}-\frac{y^{2}}{3a^{2}}=1, \\ y=x+m,\end{array}\right.$, we get $2x^{2}-2mx-m^{2}-3a^{2}=0$. Clearly, this equation has two distinct real roots.
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}+x_{2}=m, \quad x_{1} x_{2}=\frac{-m^{2}-3a^{2}}{2}.
$$

Since $\overrightarrow{A P}=3 \overrightarrow{P B}$, we have
$$
\begin{array}{c}
\left(-x_{1}, m-y_{1}\right)=3\left(x_{2}, y_{2}-m\right), \\
x_{1}=-3 x_{2}.
\end{array}
$$

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

Substituting into equation (1) gives
$$
-\frac{3}{4} m^{2}=\frac{-m^{2}-3a^{2}}{2} \Rightarrow m^{2}=6a^{2}.
$$

Noting that
$$
\begin{array}{l}
\overrightarrow{O A} \cdot \overrightarrow{O B}=x_{1} x_{2}+y_{1} y_{2} \\
=x_{1} x_{2}+\left(x_{1}+m\right)\left(x_{2}+m\right) \\
=2 x_{1} x_{2}+m\left(x_{1}+x_{2}\right)+m^{2} \\
=m^{2}-3a^{2}=3a^{2}=3.
\end{array}
$$

Therefore, $a^{2}=1$.
Thus, the equation of the hyperbola $C$ is $x^{2}-\frac{y^{2}}{3}=1$.
(2) Assume point $M$ exists, and let $M(t, 0)$.

From (1), we know that the right focus of the hyperbola is $\dot{F}(2,0)$.
Let $Q\left(x_{0}, y_{0}\right)\left(x_{0} \geqslant 1\right)$ be a point on the right branch of the hyperbola $C$.
When $x_{0} \neq 2$,
$\tan \angle Q F M=-k_{Q F}=-\frac{y_{0}}{x_{0}-2}$,
$\tan \angle Q M F=k_{Q M}=\frac{y_{0}}{x_{0}-t}$.
Since $\angle Q F M=2 \angle Q M F$, we have
$$
-\frac{y_{0}}{x_{0}-2}=\frac{2 \times \frac{y_{0}}{x_{0}-t}}{1-\left(\frac{y_{0}}{x_{0}-t}\right)^{2}}.
$$

Substituting $y_{0}^{2}=3x_{0}^{2}-3$ and simplifying, we get
$$
(4+4t)x_{0}=t^{2}+4t+3.
$$

Thus, $\left\{\begin{array}{l}4+4t=0, \\ t^{2}+4t+3=0\end{array} \Rightarrow t=-1\right.$.
When $x_{0}=2$, $\angle Q F M=90^{\circ}$, and when $t=-1$, $\angle Q M F=45^{\circ}$, satisfying
$$
\angle Q F M=2 \angle Q M F.
$$

Therefore, the point $M(-1,0)$ that satisfies the condition exists.