6. If $\triangle A B C$ satisfies:
$$
\tan \frac{A}{2} \cdot \tan \frac{C}{2}=\frac{1}{3}, b=\frac{4}{3} a,
$$

then $\sin B=$
$\begin{array}{l}\text { 6. } \frac{4}{5} \text {. } \\ \text { From } \tan \frac{A}{2} \cdot \tan \frac{C}{2}=\frac{1}{3} \\ \Rightarrow \frac{1-\cos A}{\sin A} \cdot \frac{\sin C}{1+\cos C}=\frac{1}{3} \\ \Rightarrow \frac{a^{2}-(b-c)^{2}}{(a+b)^{2}-c^{2}}=\frac{1}{3} \\ \Rightarrow \frac{a-b+c}{a+b+c}=\frac{1}{3} \\ \Rightarrow a+c=2 b \Rightarrow c=\frac{5}{3} a \\ \Rightarrow \angle C=\frac{\pi}{2} \Rightarrow \sin B=\frac{b}{c}=\frac{4}{5} .\end{array}$