Five. (Full marks 20 points) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>b>0)$ with eccentricity $e=2+\sqrt{6}-\sqrt{3}-\sqrt{2}$, a line $l$ passing through its right focus $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. Find the size of $\angle A F_{1} F_{2}$.
$$
\begin{array}{l} 
\text { Five, as shown in the figure, let } \angle A F_{1} F_{2} \\
=\alpha, \angle F_{1} A_{2} F_{2}=\beta .
\end{array}
$$

Five, as shown in the figure, let $\angle A F_{1} F_{2}$

then $\alpha+\beta=90^{\circ}$.
By the Law of Sines, we have
$$
\begin{array}{l}
\frac{\left|A F_{2}\right|}{\sin \alpha}=\frac{\left|A F_{1}\right|}{\sin 90^{\circ}}= \\
\frac{\left|F_{1} F_{2}\right|}{\sin \beta} .
\end{array}
$$

By the property of ratios, we get
$$
\begin{array}{l}
\frac{2 c}{\sin \beta}=\frac{\left|A F_{1}\right|-\left|A F_{2}\right|}{\sin 90^{\circ}-\sin \alpha}=\frac{2 a}{1-\sin \alpha} . \\
\therefore \frac{\sin \beta}{1-\sin \alpha}=\frac{c}{a}=e, \text { i.e., } \frac{\cos \alpha}{1-\sin \alpha}=e .
\end{array}
$$

Thus, $\frac{1}{\sqrt{1+\operatorname{tg}^{2} \alpha}-\operatorname{tg} \alpha}=e$, solving for $\operatorname{tg} \alpha=\frac{e^{2}-1}{2 e}$,
i.e.,
$$
\begin{array}{l}
\operatorname{tg} \alpha=\frac{(2+\sqrt{6}-\sqrt{3}-\sqrt{2})^{2}-1}{2(2+\sqrt{6}-\sqrt{3}-\sqrt{2})} \\
=\frac{(3+\sqrt{6}-\sqrt{3}-\sqrt{2})(1+\sqrt{6}-\sqrt{3}-\sqrt{2})}{2(2+\sqrt{6}-\sqrt{3}-\sqrt{2})} \\
=\frac{[\sqrt{3}(\sqrt{3}-1)+\sqrt{2}(\sqrt{3}-1)](\sqrt{3}-1)(\sqrt{2}-1)}{2(\sqrt{2}(\sqrt{2}-1)+\sqrt{3}(\sqrt{2}-1)]} \\
=\frac{(\sqrt{3}-1)^{2}}{2}=2-\sqrt{3} . \\
\therefore \alpha=15^{\circ}, \text { i.e., } \angle A F_{1} F_{2}=15^{\circ} .
\end{array}
$$