(13) If the coordinates of the two foci of a hyperbola are $F_{1}(-2,0)$ and $F_{2}(2,0)$, and a tangent line to the hyperbola intersects the $x$-axis at $Q\left(\frac{1}{2}, 0\right)$, with a slope of 2.
(1) Find the equation of the hyperbola;
(2) Let the point of tangency of the line with the hyperbola be $P$. Prove that $\angle F_{1} P Q=\angle F_{2} P Q$.
(13 Method 1 Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since it is tangent to the line $y=2\left(x-\frac{1}{2}\right)$, i.e., $y=2 x-1$, the system of equations
$$
\left\{\begin{array}{l}
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1, \\
y=2 x-1,
\end{array}\right.
$$

has only one unique solution. Therefore, the equation about $x$
$$
\frac{x^{2}}{a^{2}}-\frac{(2 x-1)^{2}}{b^{2}}=1
$$

has two equal real roots, and its discriminant is zero, so we have
$$
\left(\frac{4}{b^{2}}\right)^{2}-4\left(\frac{1}{a^{2}}-\frac{4}{b^{2}}\right)\left(-\frac{1}{b^{2}}-1\right)=0 .
$$

Simplifying, we get $b^{2}-4 a^{2}+1=0$.
Note that the coordinates of the two foci of the hyperbola are $(-2,0)$ and $(2,0)$, so its semi-focal distance is $c=2$. Thus, $a^{2}+b^{2}=4$. Solving this with the previous equation, we get $a^{2}=1, b^{2}=3$. Therefore, the equation of the hyperbola is $x^{2}-\frac{y^{2}}{3}=1$.

Since $a^{2}=1, b^{2}=3$, the equation about $x$ becomes $x^{2}-\frac{(2 x-1)^{2}}{3}=1$, from which we solve $x=2$, and substituting into $y=2 x-1$ gives $y=3$. This shows that the coordinates of the tangent point $P$ are $(2,3)$.
Thus, the slope of $F_{1} P$ is
$$
k=\frac{3-0}{2-(-2)}=\frac{3}{4},
$$

and
$$
\tan \angle F_{1} P Q=\frac{t-k}{1+k t}=\frac{1}{2},
$$

where $t=2$ is the slope of the tangent line $P Q$. Note that $F_{2}$ and $P$ have the same x-coordinate, so $F_{2} P$ is parallel to the $y$-axis, $\tan \angle F_{2} P Q=\frac{1}{t}=\frac{1}{2}$. Therefore, $\tan \angle F_{1} P Q=\tan \angle F_{2} P Q, \angle F_{1} P Q=\angle F_{2} P Q$.

Method 2 Let the equation of the hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Since the semi-focal distance $c=2$, we know $a^{2}+b^{2}=4$.
Suppose $P\left(x_{0}, y_{0}\right)$, then the equation of the tangent line at point $P$ is
$$
\frac{x_{0} x}{a^{2}}-\frac{y_{0} y}{b^{2}}=1 \text {. }
$$

Comparing this with the given tangent line equation $y=2\left(x-\frac{1}{2}\right)$, i.e., $2 x-y=1$, we get
$$
x_{0}=2 a^{2}, y_{0}=b^{2} \text {. }
$$

Substituting these into $\frac{x_{0}^{2}}{a^{2}}-\frac{y_{0}^{2}}{b^{2}}=1$ gives
$$
4 a^{2}-b^{2}=1 .
$$

Solving this with $a^{2}+b^{2}=4$, we get $a^{2}=1, b^{2}=3$. Therefore, the equation of the hyperbola is $x^{2}-$ $\frac{y^{2}}{3}=1$.

Furthermore, the coordinates of the tangent point are $\left(x_{0}, y_{0}\right)=\left(2 a^{2}, b^{2}\right)=(2,3)$, and the rest is the same as Method 1.