3. In $\triangle A B C$, $I$ is the incenter of $\triangle A B C$. If $A C+A I=B C, A B+B I=A C$, then $\angle B=$ $\qquad$
3. $\frac{2 \pi}{7}$.

As shown in Figure 1, take a point $D$ on the extension of $C A$ such that $A D=A I$. Then
$$
\begin{array}{l}
C D=C A+A D \\
=C A+A I=B C .
\end{array}
$$

By the given conditions,
$$
\begin{array}{l}
\angle C D I=\angle C B I \\
=\angle A B I .
\end{array}
$$

Since $A D=A I$, we have
$$
\angle C A I=2 \angle A D I=2 \angle A B I=\angle A B C \text {. }
$$

Thus, $\angle C A B=2 \angle A B C$.
Similarly, $\angle A B C=2 \angle A C B$.
Then $\pi=\angle A B C+\angle C A B+\angle A C B=\frac{7}{2} \angle A B C$
$$
\Rightarrow \angle A B C=\frac{2 \pi}{7} \text {. }
$$