7. Given $\alpha, \beta \in[0, \pi]$, then the maximum value of $(\sin \alpha+\sin (\alpha+\beta)) \cdot \sin \beta$ is
7. $\frac{8 \sqrt{3}}{9}$.

According to the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
(\sin \alpha + \sin (\alpha + \beta))^{2} & = (\sin \alpha + \sin \alpha \cos \beta + \cos \alpha \sin \beta)^{2} = (\sin \alpha (1 + \cos \beta) + \cos \alpha \sin \beta)^{2} \\
& \leqslant \left(\sin ^{2} \alpha + \cos ^{2} \alpha\right)\left((1 + \cos \beta)^{2} + \sin ^{2} \beta\right) = 2 + 2 \cos \beta = 4 \cos ^{2} \frac{\beta}{2}.
\end{aligned}
$$

Given $\beta \in [0, \pi]$, we have $\frac{\beta}{2} \in \left[0, \frac{\pi}{2}\right]$, thus
$$
\begin{aligned}
(\sin \alpha + \sin (\alpha + \beta)) \cdot \sin \beta & \leqslant 2 \sin \beta \cos \frac{\beta}{2} = 4 \sin \frac{\beta}{2} \cos ^{2} \frac{\beta}{2} = 8 \sqrt{\sin ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2} \cdot \frac{1}{2} \cos ^{2} \frac{\beta}{2}} \\
& \leqslant 8 \sqrt{\left(\frac{\sin ^{2} \frac{\beta}{2} + \frac{1}{2} \cos ^{2} \frac{\beta}{2} + \frac{1}{2} \cos ^{2} \frac{\beta}{2}}{3}\right)^{3}} = \frac{8 \sqrt{3}}{9}.
\end{aligned}
$$

In the above two inequalities, equality holds if and only if $\frac{\sin \alpha}{1 + \cos \beta} = \frac{\cos \alpha}{\sin \beta}$ and $\sin ^{2} \frac{\beta}{2} = \frac{1}{2} \cos ^{2} \frac{\beta}{2}$. In this case, $\beta = 2 \arctan \frac{\sqrt{2}}{2}$, $\alpha = \arctan \sqrt{2}$. Therefore, the maximum value of $(\sin \alpha + \sin (\alpha + \beta)) \cdot \sin \beta$ is $\frac{8 \sqrt{3}}{9}$.