8. In $\triangle A B C$, $a, b, c$ are the sides opposite to angles $A, B, C$ respectively. Given $a+c=2 b, A-C=\frac{\pi}{3}, \sin B=$ $\qquad$ .
\begin{array}{l}\text { 8. } \frac{\sqrt{39}}{8} \text {. } \\ a+c=2 b \Rightarrow 2 R \sin A+2 R \sin C=2 \cdot 2 R \sin B \Rightarrow \sin A+\sin C=2 \sin B \Rightarrow \\ 2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}=2 \sin B \Rightarrow 2 \sin \frac{180^{\circ}-B}{2} \cos \frac{A-C}{2}=2 \sin B \Rightarrow 2 \cos \frac{B}{2} \cos \frac{\pi}{6}=2 \cdot \\ 2 \sin \frac{B}{2} \cos \frac{B}{2}, 0<B<180^{\circ} \Rightarrow \cos \frac{B}{2} \neq 0 \Rightarrow \sin \frac{B}{2}=\frac{1}{2} \cos \frac{\pi}{6}=\frac{\sqrt{3}}{4} \Rightarrow \cos \frac{B}{2}=\sqrt{1-\sin ^{2} \frac{B}{2}} \\ =\frac{\sqrt{13}}{4} \text {, so } \sin B=2 \sin \frac{B}{2} \cos \frac{B}{2}=\frac{\sqrt{39}}{8} .\end{array}