$3.45^{\circ}$.
As shown in Figure 4, connect $O C$. Since $P C$ is tangent to $\odot O$ at point $C$, then $O C$ $\perp P C$.

Since $O A=$ $O C$, therefore,
$$\begin{array}{l} \angle O A C=\angle O C A=\frac{1}{2} \angle P O C . \\ \text { Also } \angle A P Q=\angle C P Q=\frac{1}{2} \angle A P C \text {, so } \\ \angle P Q C=\angle P A Q+\angle A P Q \\ =\frac{1}{2}(\angle P O C+\angle A P C)=\frac{1}{2} \times 90^{\circ}=45^{\circ} . \end{array}$$