2. In $\triangle A B C$, $\angle A=45^{\circ}, \angle B=30^{\circ}$, $A D$ is the median of $\triangle A B C$. Then $\angle A D C=$ $\qquad$
$2.45^{\circ}$.
As shown in Figure 2, draw $C E \perp A B$ at point $E$, and connect $D E$. Then
$$
\begin{array}{l}
D E=\frac{1}{2} B C \\
=B D=D C, \\
A E=E C, \\
\angle D C E=60^{\circ} .
\end{array}
$$

Therefore, $\triangle D C E$ is an equilateral triangle
$$
\Rightarrow D E=E C, \angle D E C=60^{\circ} \text {. }
$$

Thus, $\angle A E D=90^{\circ}+60^{\circ}=150^{\circ}$.
Hence $A E=D E$
$$
\Rightarrow \angle E A D=\angle E D A=\frac{180^{\circ}-\angle A E D}{2}=15^{\circ} \text {. }
$$

In $\triangle A B D$,
$$
\angle A D C=\angle B+\angle E A D=30^{\circ}+15^{\circ}=45^{\circ}
$$