Example 12 (Question from the 11th "Hope Cup" Invitational Competition) As shown in Figure 7-11, the ratio of the side lengths of the upper and lower bases of the frustum of a triangular pyramid $A B C-D E F$ is $1: 2, G$ is the midpoint of $C F$, then the ratio of the volumes of the upper and lower parts of the frustum divided by the section $A G E$ is $\qquad$ (the volume of the frustum $V=\frac{h}{3}(t+\sqrt{f F}+F)$, where $h$ is the height of the frustum, $f, F$ are the areas of the upper and lower bases).
Solve: Fill in $\frac{2}{5}$. Reason: Let $H, I$ be the midpoints of $AD, BE$ respectively, and let $HI$ intersect $AE$ at $J$. Since the ratio of the side lengths of the upper and lower bases of the frustum is $1:2$, we can assume the areas of the upper and lower bases are $4S$ and $16S$, respectively. Thus, the area of the middle section $HIG$ is $9S$.
Also, $HJ:JI=2:1$, so $S_{\triangle HJG}=6S, S_{\triangle JIG}=3S$.
Let the height of the frustum be $2h$, then $V_{\text{frustum ABC-DEF}}=\frac{1}{3} \cdot 2h(4S+16S+\sqrt{4 \cdot 16S^2})=\frac{56}{3}hS$.
$$
\begin{aligned}
V_{\text{part F of the frustum}} & =V_{\text{pyramid A-HJG}}+\left(V_{\text{frustum HKG-DEF}}-V_{\text{pyramid E-JIG}}\right) \\
& =\frac{1}{3}h \cdot 6S+\left[\frac{1}{3}h(9S+16S+\sqrt{9 \cdot 16S^2})-\frac{1}{3}h \cdot 3S\right] \\
& =2hS+\frac{34}{3}hS=\frac{40}{3}hS .
\end{aligned}
$$