5. In $\triangle A B C$, suppose
$$
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A},
$$

and $\cos (A-B)+\cos C=1-\cos 2 C$. Then the value of $\frac{a+c}{b}$ is
5. $\frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}}$.

The first equation, derived from the Law of Sines, is
$$
\begin{array}{l}
\frac{a+b}{a}=\frac{\sin B}{\sin B-\sin A}=\frac{b}{b-a} \\
\Rightarrow b^{2}-a^{2}=a b \Rightarrow\left(\frac{a}{b}\right)^{2}+\left(\frac{a}{b}\right)-1=0 .
\end{array}
$$

Solving the equation and taking the positive value, we get $\frac{a}{b}=\frac{\sqrt{5}-1}{2}$.
From the second equation, we have
$$
\begin{array}{l}
\cos (A-B)-\cos (A+B)=2 \sin ^{2} C \\
\Rightarrow \sin A \cdot \sin B=\sin ^{2} C \Rightarrow a b=c^{2} \\
\Rightarrow\left(\frac{c}{b}\right)^{2}=\frac{a}{b}=\frac{\sqrt{5}-1}{2} \Rightarrow \frac{c}{b}=\sqrt{\frac{\sqrt{5}-1}{2}} \\
\Rightarrow \frac{a+c}{b}=\frac{a}{b}+\frac{c}{b}=\frac{\sqrt{5}-1}{2}+\sqrt{\frac{\sqrt{5}-1}{2}} .
\end{array}
$$