1. Let the side length of the regular hexagon $A B C D E F$ be 1. Then $(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E})=$ $\qquad$ .
$$
-1 .-3 \text {. }
$$

As shown in Figure 1, establish a Cartesian coordinate system, and set $C(1,0)$.
Thus, $B\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right), A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$,
$$
\begin{array}{l}
D\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right), E\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) . \\
\text { Then } \overrightarrow{A B}+\overrightarrow{D C}=(1,0)+\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)=\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right), \\
\overrightarrow{A D}+\overrightarrow{B E}=(1,-\sqrt{3})+(-1,-\sqrt{3}) \\
=(0,-2 \sqrt{3}) .
\end{array}
$$

Therefore, $(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E})$
$$
=\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right) \cdot(0,-2 \sqrt{3})=-3 .
$$