2. In $\triangle A B C$, $A B=5, A C=12, C B=$ $13, D, E$ are points on side $B C$, satisfying $B D=1, C E=$ 8. Then the degree of $\angle D A E$ is
$$
\begin{array}{l} 
\mathbf{2 . 4 5} . \\
\text { As shown in Figure } 5, \text { let } \angle A D C \\
=\alpha, \angle A E B=\beta .
\end{array}
$$

As shown in Figure 5, let $\angle A D C$
Since $A B^{2}+A C^{2}$
$$
=5^{2}+12^{2}=13^{2}=C B^{2} \text {. }
$$

Therefore, $\angle B A C=90^{\circ}$.
From the given conditions, $B A=B E, C A=C D$.
Thus, $\angle B A E=\beta, \angle D A C=\alpha$.
Since $\angle B A C=90^{\circ}$, we have
$$
\begin{array}{l}
\alpha+\beta-\angle D A E=90^{\circ} . \\
Also, \alpha+\beta+\angle D A E=180^{\circ} \\
\Rightarrow 2 \angle D A E=90^{\circ} \\
\Rightarrow \angle D A E=45^{\circ} .
\end{array}
$$