1. In Rt $\triangle A B C$, $\angle C=90^{\circ}, A B=1$, point $E$ is the midpoint of side $A B$, and $C D$ is the altitude on side $A B$. Then the maximum value of $(\overrightarrow{C A} \cdot \overrightarrow{C D}) \cdot(\overrightarrow{C A} \cdot \overrightarrow{C E})$ is $\qquad$ (Li Weiwei provided the problem)
1. $\frac{2}{27}$.

Let $C A=b, C B=a$, then $a^{2}+b^{2}=A B^{2}=1, C D=2 \frac{S_{\triangle A B C}}{A B}=a b$, thus
$$
\begin{aligned}
(\overrightarrow{C A} \cdot \overrightarrow{C D}) \cdot(\overrightarrow{C A} \cdot \overrightarrow{C E}) & =|\overrightarrow{C D}|^{2} \cdot \frac{1}{2} \overrightarrow{C A} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=\frac{1}{2}|\overrightarrow{C D}|^{2} \cdot|\overrightarrow{C A}| \\
& =\frac{1}{2} a^{2} b^{4}=2\left(1-b^{2}\right) \cdot \frac{b^{2}}{2} \cdot \frac{b^{2}}{2} \\
& \leqslant\left(\frac{(1-b)^{2}+\frac{b^{2}}{2}+\frac{b^{2}}{2}}{3}\right)^{3}=\frac{2}{27} .
\end{aligned}
$$

The equality holds if and only if $b^{2}=\frac{2}{3}$, i.e., $a=\frac{\sqrt{3}}{3}, b=\frac{2 \sqrt{3}}{3}$. Therefore, the maximum value of $(\overrightarrow{C A} \cdot \overrightarrow{C D}) \cdot(\overrightarrow{C A} \cdot \overrightarrow{C E})$ is $\frac{2}{27}$.