1. Let the side length of the regular hexagon $A B C D E F$ be 1, then $(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E})=$
$$
\begin{array}{l}
\text { As shown in the figure, }(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E}) \\
=\overrightarrow{A B} \cdot \overrightarrow{A D}+\overrightarrow{A B} \cdot \overrightarrow{B E}+\overrightarrow{D C} \cdot \overrightarrow{A D}+\overrightarrow{D C} \cdot \overrightarrow{B E} \\
=2 \cos 60^{\circ}-2 \cos 60^{\circ}-2 \cos 60^{\circ}-2=-3 .
\end{array}
$$

Therefore, $(\overrightarrow{A B}+\overrightarrow{D C}) \cdot(\overrightarrow{A D}+\overrightarrow{B E})=-3$.