65. Find the maximum value of the real number $\lambda$ such that point $P$ is inside the acute $\triangle ABC$, $\angle PAB = \angle PBC = \angle PCA$, and the rays $AP$, $BP$, $CP$ intersect the circumcircles of $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ at $A_1$, $B_1$, $C_1$, respectively, so that $S_{\triangle A_1BC} + S_{\triangle B_1CA} + S_{\triangle C_1AB} \geqslant \lambda S_{\triangle ABC}$.
65. Let $A A_{1} \cap B C=A_{2}$. From $\angle P A B=\angle P B C=\angle P A_{1} C$, we know $A B \parallel A_{1} C$. Therefore, $\triangle A_{1} A_{2} C \sim \triangle A A_{2} B$
$$\frac{S_{\triangle A_{1} B C}}{S_{\triangle A B C}}=\frac{A A_{2}}{A A_{2}}=\frac{C A_{2}}{B A_{2}}$$

Similarly, let $B_{2}=B B_{1} \cap C A, C_{2}=C C_{1} \cap A B$, then
$$\begin{array}{l}
\frac{S_{\triangle B_{1} C A}}{S_{\triangle A B C}}=\frac{A B_{2}}{C B_{2}} \\
\frac{S_{\triangle C_{1} A B}}{S_{\triangle A B C}}=\frac{B C_{2}}{C_{2} A}
\end{array}$$

Thus, by the AM-GM inequality and Ceva's theorem, we have
$$\begin{array}{l}
\frac{S_{\triangle A_{1} B C}+S_{\triangle B_{1} C A}+S_{\triangle C_{1} A B}}{S_{\triangle A B C}}=\frac{C A_{2}}{B A_{2}}+\frac{A B_{2}}{C B_{2}}+\frac{B C_{2}}{C_{2} A} \geqslant \\
3 \sqrt[3]{\frac{C A_{2}}{B A_{2}} \cdot \frac{A B_{2}}{C B_{2}} \cdot \frac{B C_{2}}{C_{2} A}}=3
\end{array}$$

When $\triangle A B C$ is an equilateral triangle and $P$ is the center, $\angle P A B=\angle P B C=\angle P C A=30^{\circ}$, and $A_{2} B=A_{2} C$, $B_{2} A=B_{2} C, C_{2} A=C_{2} B$. In this case, the equality holds.

In summary, the maximum value of $\lambda$ is 3.
Similarly, $\frac{G Q}{G C}=\frac{a-c}{c+2 a}$. Noting the area relationship $F=3 S_{\triangle G B C}$, we have
$$\frac{S_{\triangle G P Q}}{F}=\frac{1}{3} \cdot \frac{G P \cdot G Q}{G B \cdot G C}=\frac{1}{3} \cdot \frac{(b-c)(a-c)}{(b+2 c)(c+2 a)}$$