I was able to show the above by equating the solution of $cos \ 5\theta=0$ (which gives $\pi/10$) and the solution of ${16cos}^5\theta - {20cos}^3\theta+5cos\theta = 0$ (which is what you get when you use De Moivre's Theorem with binomial expansion). However, I do not understand how the second part of the question has anything to do with the first, and thus am lost as to where to begin.
What you have to realize about the equation $\cos5\theta=0$ is that $\frac{\pi}{10}$ is a solution because $\cos\frac{\pi}{2}=0$, but also that $\frac{7\pi}{10}$ is a solution because $\cos\frac{35\pi}{10}=\cos\frac{5\pi}2=\cos\left(2\pi+\frac{\pi}2\right)=\cos\frac\pi 2=0$. Therefore, we can use this equation to find both $\cos\frac\pi {10}$ and $\cos\frac{7\pi}{10}$.
We have the following: $$16\cos^5\theta-20\cos^3\theta+5\cos\theta=0$$
We want to solve this equation to find $\cos \frac{\pi}{10}$ and $\cos \frac{7\pi}{10}$, which we know is not $0$. Therefore, we can divide by $\cos \theta$: $$16\cos^4\theta-20\cos^2\theta+5=0$$
Now, we can use the quadratic formula to find $\cos^2\theta$: $$\cos^2\theta=\frac{20\pm\sqrt{20^2-4\cdot16\cdot5}}{2(16)}$$ $$\cos^2\theta=\frac{20\pm4\sqrt 5}{32}$$ $$\cos^2\theta=\frac{5\pm\sqrt5}{8}$$
Now, since $\frac{\pi}{10}$ is very close to $0$, it will have a bigger $\cos$, so we want to take the bigger solution. Therefore, the $\pm$ becomes a $+$. $$\cos^2\frac{\pi}{10}=\frac{5+\sqrt5}{8}$$ $\frac\pi{10}$ is in Quadrant I, so we take the positive square root of both sides: $$\cos\frac\pi{10}=\sqrt{\frac{5+\sqrt5}{8}}$$
Now, for $\frac{7\pi}{10}$, we have: $$\cos^2\frac{7\pi}{10}=\frac{5\pm\sqrt5}{8}$$ For this, you need to ask yourself two questions:
Good luck finding $\cos\frac{7\pi}{10}$!
