3. In $\triangle A B C$, it is known that
$$
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A},
$$

and $\cos (A-B)+\cos C=1-\cos 2 C$.
Then $\frac{a+c}{b}=$
$$
\begin{array}{l}
\text { 3. } \frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}} . \\
\text { From } \cos (A-B)+\cos C=1-\cos 2 C \\
\Rightarrow \cos (A-B)-\cos (A+B)=2 \sin ^{2} C \\
\Rightarrow 2 \sin A \cdot \sin B=2 \sin ^{2} C \Rightarrow a b=c^{2} . \\
\text { From } \frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A} \Rightarrow b^{2}-a^{2}=a b \\
\Rightarrow \frac{a}{b}=\frac{\sqrt{5}-1}{2} .
\end{array}
$$

Therefore, $\frac{c}{b}=\sqrt{\frac{\sqrt{5}-1}{2}}$.