A = sin²α + sin²(α - π/5) + sin²(α + π/3)

To simplify the expression \( A = \sin^2 \alpha + \sin^2\left(\alpha - \frac{\pi}{5}\right) + \sin^2\left(\alpha + \frac{\pi}{3}\right) \), we can use the identity:

\[
\sin^2 x = \frac{1 - \cos(2x)}{2}
\]

Applying this, we can rewrite each term:

1. \( \sin^2 \alpha = \frac{1 - \cos(2\alpha)}{2} \)
2. \( \sin^2\left(\alpha - \frac{\pi}{5}\right) = \frac{1 - \cos\left(2\left(\alpha - \frac{\pi}{5}\right)\right)}{2} = \frac{1 - \cos(2\alpha - \frac{2\pi}{5})}{2} \)
3. \( \sin^2\left(\alpha + \frac{\pi}{3}\right) = \frac{1 - \cos\left(2\left(\alpha + \frac{\pi}{3}\right)\right)}{2} = \frac{1 - \cos(2\alpha + \frac{2\pi}{3})}{2} \)

Putting these together, the expression for \( A \) becomes:

\[
A = \frac{1 - \cos(2\alpha)}{2} + \frac{1 - \cos(2\alpha - \frac{2\pi}{5})}{2} + \frac{1 - \cos(2\alpha + \frac{2\pi}{3})}{2}
\]

Combining the fractions:

\[
A = \frac{3}{2} - \frac{1}{2}\left(\cos(2\alpha) + \cos(2\alpha - \frac{2\pi}{5}) + \cos(2\alpha + \frac{2\pi}{3})\right)
\]

This can be further simplified depending on the evaluations of the cosine terms.

If needed, numerical or graphical methods can give insight into specific values of \( A \) for certain \(\alpha\).