13.5.3 * Given: The hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$ has an eccentricity $e=2+\sqrt{6}-\sqrt{3}-\sqrt{2}$, and a line $e$ passing through its right focus $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. Find the value of $\angle A F_{1} F_{2}$.
Let $\angle A F_{1} F_{2}=\alpha, \angle F_{1} A F_{2}=\beta$, then $\alpha + \beta = 90^{\circ}$. By the Law of Sines, we have $\frac{\left|A F_{2}\right|}{\sin \alpha}=\frac{\left|A F_{1}\right|}{\sin 90^{\circ}}=\frac{\left|F_{1} F_{2}\right|}{\sin \beta}$, so $\frac{2 c}{\sin \beta}=\frac{\left|A F_{1}\right|-\left|A F_{2}\right|}{\sin 90^{\circ}-\sin \alpha}=\frac{2 a}{1-\sin \alpha}$, thus $\frac{\sin \beta}{1-\sin \alpha}=\frac{c}{a}=e$, which means $\frac{\cos \alpha}{1-\sin \alpha}=e$.

This implies $\frac{1}{\sqrt{1+\tan ^{2} \alpha-\tan \alpha}}=e$, leading to $\tan \alpha=\frac{e^{2}-1}{2 e}$. Therefore,
$$
\begin{aligned}
\tan \alpha & =\frac{(2+\sqrt{6}-\sqrt{3}-\sqrt{2})^{2}-1}{2(2+\sqrt{6}-\sqrt{3}-\sqrt{2})} \\
& =\frac{(3+\sqrt{6}-\sqrt{3}-\sqrt{2})(1+\sqrt{6}-\sqrt{3}-\sqrt{2})}{2(2+\sqrt{6}-\sqrt{3}-\sqrt{2})} \\
& =\frac{(\sqrt{3}-1)^{2}}{2}=2-\sqrt{3} .
\end{aligned}
$$

Thus, $\alpha=15^{\circ}$, i.e., $\angle A F_{1} F_{2}=15^{\circ}$.