5. As shown in Figure 1, quadrilateral $A C D B$ and parallelogram $A E D F$ satisfy:
$$
\begin{array}{l}
6 \overrightarrow{A E}+3 \overrightarrow{D E}=2 \overrightarrow{A C}, 4 \overrightarrow{F B}+5 \overrightarrow{F D}=2 \overrightarrow{A D} . \\
\text { Then } \frac{S_{\text {quadrilateral } A C D B}}{S_{\text {quadrilateral } A E D F}}=
\end{array}
$$
5. $\frac{27}{8}$.

From the given equations, we have
$$
\begin{array}{l}
4 \overrightarrow{A E}+3 \overrightarrow{D E}+2 \overrightarrow{C E}=\mathbf{0}, \\
4 \overrightarrow{F B}+3 \overrightarrow{F D}+2 \overrightarrow{F A}=\mathbf{0} .
\end{array}
$$

By the Mercedes-Benz theorem, we get
$$
\begin{array}{l}
S_{\triangle E A D}: S_{\triangle E D C}: S_{\triangle E C A}=2: 4: 3, \\
S_{\triangle F A D}: S_{\triangle F D B}: S_{\triangle F B A}=4: 2: 3 .
\end{array}
$$

Since quadrilateral $A E D F$ is a parallelogram, we have
$$
\begin{array}{l}
S_{\triangle E A D}=S_{\triangle E A D} . \\
\text { Therefore, } \frac{S_{\text {quadrilateral } A C D B}}{S_{\text {quadrilateral } A E D F}}=\frac{2+3+4+4+6+8}{4+4}=\frac{27}{8} .
\end{array}
$$