Given (alpha + beta) = 1 and (alpha − beta) = 1/2, with alpha, beta ext in [0, extpi/2], find (alpha) and (beta).

We are given $\sin(\alpha + \beta) = 1$.
Since $\alpha, \beta \in \left[0, \frac{\pi}{2}\right]$, it follows that $\alpha + \beta \in [0, \pi]$.
The only angle in this interval for which $\sin x = 1$ is $x = \frac{\pi}{2}$.
So, we have:

$$\alpha + \beta = \frac{\pi}{2} \quad (*)$$

We are given $\sin(\alpha - \beta) = \frac{1}{2}$.
Since $\alpha, \beta \in \left[0, \frac{\pi}{2}\right]$, it follows that $\alpha - \beta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
The only angle in this interval for which $\sin x = \frac{1}{2}$ is $x = \frac{\pi}{6}$.
So, we have:

$$\alpha - \beta = \frac{\pi}{6} \quad (**)$$

Add equations $(*)$ and $(**)$:

$$(\alpha + \beta) + (\alpha - \beta) = \frac{\pi}{2} + \frac{\pi}{6}$$ $$2\alpha = \frac{3\pi}{6} + \frac{\pi}{6}$$ $$2\alpha = \frac{4\pi}{6}$$ $$2\alpha = \frac{2\pi}{3}$$ $$\alpha = \frac{\pi}{3}$$

Substitute $\alpha = \frac{\pi}{3}$ into equation $(*)$:

$$\frac{\pi}{3} + \beta = \frac{\pi}{2}$$ $$\beta = \frac{\pi}{2} - \frac{\pi}{3}$$ $$\beta = \frac{3\pi}{6} - \frac{2\pi}{6}$$ $$\beta = \frac{\pi}{6}$$

Both $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{6}$ are within the given domain $\left[0, \frac{\pi}{2}\right]$.

First argument: $\alpha + 2\beta$

$$\alpha + 2\beta = \frac{\pi}{3} + 2\left(\frac{\pi}{6}\right)$$ $$\alpha + 2\beta = \frac{\pi}{3} + \frac{\pi}{3}$$ $$\alpha + 2\beta = \frac{2\pi}{3}$$

Second argument: $2\alpha - \beta$

$$2\alpha - \beta = 2\left(\frac{\pi}{3}\right) - \frac{\pi}{6}$$ $$2\alpha - \beta = \frac{2\pi}{3} - \frac{\pi}{6}$$ $$2\alpha - \beta = \frac{4\pi}{6} - \frac{\pi}{6}$$ $$2\alpha - \beta = \frac{3\pi}{6}$$ $$2\alpha - \beta = \frac{\pi}{2}$$

Evaluate $\tan(\alpha + 2\beta)$:

$$\tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$$

Evaluate $\tan(2\alpha - \beta)$:

$$\tan\left(\frac{\pi}{2}\right)$$

The tangent function is undefined at $\frac{\pi}{2}$.

Since one of the terms in the product is undefined, the entire product is undefined.

The final answer is $\text{Undefined}$.