4.79 Method I. Let $A=\cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5}$. Multiplying both sides of this equation by $2 \sin \frac{3 \pi}{5}$, we get

$2 A \sin \frac{3 \pi}{5}=2 \sin \frac{3 \pi}{5} \cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5} ; 2 A \sin \frac{3 \pi}{5}=\sin \frac{6 \pi}{5} \cos \frac{6 \pi}{5} ;$

$2 A \sin \frac{3 \pi}{5}=\frac{1}{2} \sin \frac{12 \pi}{5} ; 2 A \sin \frac{3 \pi}{5}=\frac{1}{2} \sin \frac{2 \pi}{5}$.

But $\sin \frac{2 \pi}{5}=\sin \left(\pi-\frac{3 \pi}{5}\right)=\sin \frac{3 \pi}{5}$, from which $A=\frac{1}{4}$.

Method II. We have

$$\begin{aligned} & A=\cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5}=\frac{2 \sin \frac{3 \pi}{5} \cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5}}{2 \sin \frac{3 \pi}{5}}= \\ & =\frac{\sin \frac{12 \pi}{5}}{4 \sin \frac{3 \pi}{5}}=\frac{\sin \frac{2 \pi}{5}}{4 \sin \frac{3 \pi}{5}}=\frac{1}{4} \end{aligned}$$