1. (20 points) As shown in the figure, $\angle A B E=\angle D C F=90^{\circ}, A B=3, D C=5, B C=6, B E=E F=F C, A F$ intersects $D E$ at $G$. Then the sum of the areas of triangle $D F G$ and triangle $A G E$ is $\qquad$ .
【Analysis】Draw $E H \perp B C$ intersecting $A F$ at $H$, and draw $F I \perp B C$ intersecting $D E$ at $I$. Solve based on the ratio of the areas of triangles with equal heights being equal to the ratio of their bases.

【Solution】Solution: Draw $E H \perp B C$ intersecting $A F$ at $H$, and draw $F I \perp B C$ intersecting $D E$ at $I$. Since $\angle A B E=\angle D C F=90^{\circ}, A B=3, D C=5, B C=6, B E=E F=F C$,

Therefore, $E H=1.5, F I=2.5$, the area of triangle $E F H$ = the area of triangle $A E H$ = $2 \times 3 \div 2 \div 2=1.5$, the area of triangle $E F I$ = the area of triangle $F I D$ = $2 \times 5 \div 2 \div 2=2.5$.

Therefore, $H G: G F=1.5: 2.5, I G: G E=2.5: 1.5$,
So the area of triangle $E G H$ = $1.5 \times 1.5 \div(1.5+2.5)=\frac{9}{16}$,
The area of triangle $G F I$ = $2.5 \times 2.5 \div(1.5+2.5)=\frac{25}{16}$.
Thus, the sum of the areas of triangle $D F G$ and triangle $A G E$ = the area of triangle $A E H$ + the area of triangle $E G H$ + the area of triangle $F I D$
+ the area of triangle $G F I$,
$$
=1.5+\frac{9}{16}+\frac{25}{16}+2.5=\frac{49}{8} \text {. }
$$

Therefore, the answer is: $\frac{49}{8}$.