$\begin{aligned} & \text { If } \sin (\alpha+\beta)=1, \sin (\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in\left[0, \frac{\pi}{2}\right] \\ & \text { then } \tan (\alpha+2 \beta) \cdot \tan (2 \alpha+\beta)=\end{aligned}$

Given: $\sin(\alpha+\beta)=1$ and $\sin(\alpha-\beta)=\frac{1}{2}$, with $\alpha, \beta \in \left[0, \frac{\pi}{2}\right]$.

From $\sin(\alpha+\beta)=1$, and since $\alpha+\beta \in [0, \pi]$, the only angle satisfying this is $\alpha+\beta = \frac{\pi}{2}$.

Similarly, $\sin(\alpha-\beta)=\frac{1}{2}$ with $\alpha-\beta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ gives $\alpha-\beta = \frac{\pi}{6}$.

Solving the system:

$\alpha+\beta = \frac{\pi}{2}$

$\alpha-\beta = \frac{\pi}{6}$

Adding yields $2\alpha = \frac{2\pi}{3}$, so $\alpha = \frac{\pi}{3}$. Substituting gives $\beta = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}$.

Now compute $\tan(\alpha+2\beta) = \tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}$ and $\tan(2\alpha+\beta) = \tan\left(\frac{5\pi}{6}\right) = -\frac{1}{\sqrt{3}}$.

The product is $(-\sqrt{3}) \cdot \left(-\frac{1}{\sqrt{3}}\right) = 1$.