2. Given that $A D$ is the median of $\triangle A B C$, $\angle A B C=$ $30^{\circ}, \angle A D C=45^{\circ}$. Then $\angle A C B=$ $\qquad$
2. $105^{\circ}$.

As shown in Figure 7, draw
the altitude $C H$. Then
$$
\begin{array}{c}
H D=D C \\
=\frac{1}{2} B C \\
=H C .
\end{array}
$$

Since $\angle H A D$
$$
\begin{array}{l} 
\quad=\angle B A D \\
= \angle A D C-\angle A B D=45^{\circ}-30^{\circ}=15^{\circ}, \\
\angle H D A=\angle H D C-\angle A D C \\
= 60^{\circ}-45^{\circ}=15^{\circ},
\end{array}
$$

Therefore, $H A=D H=H C$.
Thus, $\triangle C H A$ is an isosceles right triangle.
Hence $\angle A C H=45^{\circ}$,
$$
\angle A C B=60^{\circ}+45^{\circ}=105^{\circ} \text {. }
$$