\[\begin{array}{l} {\text{а) }}4\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5} = \frac{{2 \cdot \sin \frac{{2\pi }}{5}\cos \frac{{2\pi }}{5}}}{{\sin \frac{\pi }{5}}} = \frac{{\sin \frac{{4\pi }}{5}}}{{\sin \frac{\pi }{5}}} = \hfill \\ \left[ {\sin \alpha = \sin \left( {\pi - \alpha } \right)} \right] = \frac{{\sin \frac{\pi }{5}}}{{\sin \frac{\pi }{5}}} = 1. \hfill \\ {\text{б) }}{\left( {\cos \frac{\pi }{5} - \cos \frac{{2\pi }}{5}} \right)^2} = {\cos ^2}\frac{\pi }{5} - 2\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5} + {\cos ^2}\frac{{2\pi }}{5} = \hfill \\ \frac{1}{2}\left( {2 + \cos \frac{{2\pi }}{5} - 4\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5} - \cos \frac{\pi }{5}} \right) = \frac{1}{2}\left( {1 + \cos \frac{{2\pi }}{5} - \cos \frac{\pi }{5}} \right), \hfill \\ {\text{т}}{\text{.е}}{\text{. }}{x^2} = \frac{1}{2}\left( {1 - x} \right). \hfill \\ \end{array}\]

\[{\text{а) }}\frac{1}{4};{\text{ б) }}\frac{1}{2}\]