From the given conditions,
(As shown in Figure 7),
$$\begin{array}{l} \angle \mathrm{EAB} \\ =\angle \mathrm{PAB}, \\ \angle \mathrm{FAC} \end{array}$$
$$\begin{array}{l} =\angle \mathrm{PAC}, \text{ hence } \\ \angle \mathrm{EAP}=2 \angle \mathrm{PAB}, \angle \mathrm{FAP}=2 \angle \mathrm{PAC}. \\ \therefore \angle \mathrm{EAF}=\angle \mathrm{EAP}+\angle \mathrm{FAP}=\angle \mathrm{PAB} \\ +\angle \mathrm{PAC}=8 \angle B A C=\text{ a constant, then } \mathrm{S}_{\triangle A E F} \\ =\frac{1}{2} \mathrm{AE} \cdot \mathrm{AF} \sin \angle \mathrm{EAF}=\frac{1}{2} \mathrm{AP}^{2} \\ \end{array}$$
. $\sin 2 \angle B A C$. To minimize $\mathrm{S}_{\triangle \mathrm{AFF}}$, it is only necessary for $\mathrm{AP}$ to be the smallest, so draw $\mathrm{PP} \perp \mathrm{BC}$ through $\mathrm{A}$, with $\mathrm{P}$ being the foot of the perpendicular, and point $\mathrm{P}$ is the required point; similarly, to maximize $\mathrm{S}_{\triangle A E F}$, it is only necessary for $\mathrm{AP}$ to be the largest, so when point $\mathrm{P}$ coincides with point $\mathrm{B}$, $\mathrm{S}_{\triangle A E F}$ is the largest.