Let $A=1+\cos\frac{\pi}{5}+\cos\frac{3\pi}{5} +\cos\frac{\pi}{5}\cos\frac{3\pi}{5}$, and $\theta=\frac{\pi}{5}$. We have: $$2\theta = \pi - 3\theta \implies \cos 2\theta + \cos 3\theta = 0 \implies 4\cos^3\theta + 2\cos^2\theta - 3\cos\theta -1=0$$ Factorization leads to $$(\cos\theta+1)(4\cos^2\theta -2\cos\theta -1)=0$$ Clearly, $\cos\theta \ne 0$, hence $$4\cos^2\theta -2\cos\theta -1=0$$ Therefore $\cos\theta = \cos\frac{\pi}{5}$ is one solution to $$4x^2 - 2x -1=0$$ Similarly, we can also show $\cos\frac{3\pi}{5}$ is also a root to the above equation. Then by the Vieta's theorem, we find $$A=1+\frac{2}{4}-\frac{1}{4}=\boxed{\frac{5}{4}}$$