$• \ \text{a)}$ $\color{#3368b8}{\text{0}}$ Com conhecimento do cosseno e seno da soma, assim como seus respectivos arcos-duplos, pode-se escrever:\begin{matrix} \cos{\frac{2\pi}{5}} \cdot \cos{\frac{\pi}{10}} - \sin{\frac{2\pi}{5}}\sin{\frac{\pi}{10}} \end{matrix}$\color{orangered}{\text{Obs:}}$\begin{matrix} \cos{\frac{2\pi}{5}} = (\cos^2{\frac{\pi}{5}} - \sin^2{\frac{\pi}{5}} ) &,& \sin{\frac{2\pi}{5}} = 2\sin{\frac{\pi}{5}}\cos{\frac{\pi}{5}} \end{matrix}Sabido o cosseno da soma,\begin{matrix} \cos{\frac{2\pi}{5}} \cdot \cos{\frac{\pi}{10}} - \sin{\frac{2\pi}{5}}\sin{\frac{\pi}{10}} = \cos(\frac{2\pi}{5} + \frac{\pi}{10}) \end{matrix}Portanto,\begin{matrix} \cos(\frac{2\pi}{5} + \frac{\pi}{10}) = \cos{\frac{\pi}{2}} =0 \ \ \ \tiny{\blacksquare} \end{matrix}$• \ \text{b)}$ $\color{#3368b8}{\text{1/4}}$ Conforme resultado anterior, sabemos que:\begin{matrix} \cos{\frac{2\pi}{5}} \cdot \cos{\frac{\pi}{10}} - 2\left( \sin{\frac{\pi}{10}}\cos{\frac{\pi}{5}}\right)\sin{\frac{\pi}{5}} = 0 \end{matrix}Denotemos $\sin{\frac{\pi}{10}}\cos{\frac{\pi}{5}} = x$, logo:\begin{matrix} \cos{\frac{2\pi}{5}} \cdot \cos{\frac{\pi}{10}} =2x\sin{\frac{\pi}{5}} \end{matrix}Conforme arco-duplo, \begin{matrix} \sin{\frac{\pi}{5}} = 2\sin{\frac{\pi}{10}}\cos{\frac{\pi}{10}} \end{matrix}Consequentemente,\begin{matrix} \cos{\frac{2\pi}{5}} = 4x\sin{\frac{\pi}{10}} \end{matrix}Conhecida a relação entre complementares, sabemos que:\begin{matrix} \cos{\frac{4\pi}{10}} = \sin{(\frac{\pi}{2} - \frac{4\pi}{10} )} = \sin{\frac{\pi}{10}} \end{matrix}$\color{orangered}{\text{Obs:}}$ $ \cos{\frac{2\pi}{5}} = \cos{\frac{4\pi}{10}}$ Portanto,\begin{matrix} x = \dfrac{1}{4} \ \ \ \tiny{\blacksquare} \end{matrix}