How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.

The complex roots of $x^5-1$ are: $$ \begin{align} x_1&=1\\ x_2&=\cos\frac{2\pi}5+i\sin\frac{2\pi}5\\ x_3&=-\cos\frac{\pi}5+i\sin\frac{\pi}5\\ x_4&=-\cos\frac{\pi}5-i\sin\frac{\pi}5\\ x_5&=\cos\frac{2\pi}5-i\sin\frac{2\pi}5 \end{align} $$ using Vieta's formulas you get $$0=x_1+x_2+\dots+x_5=1+2\left(\cos\frac{2\pi}5-\cos\frac\pi5\right)=0,$$ which yields your first equation.

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From now on, let $\varphi=\frac{\pi}5$ (for brevity).

We know that $\cos2\varphi-\cos\varphi+\frac12=(2\cos^2\varphi-1)-\cos\varphi+\frac12=2\cos^2\varphi-\cos\varphi-\frac12=0$, i.e. $$2\cos^2\varphi-\cos\varphi=\frac12.$$ Now from $\cos2\varphi=\cos\varphi-\frac12$ you get $$\cos\varphi\cos2\varphi=\cos^2\varphi-\frac{\cos\varphi}2=\frac{2\cos^2\varphi-\cos\varphi}2=\frac14.$$

(Or, as suggested in Chandrasekhar's answer, from $2\cos^2\varphi-\cos\varphi=\frac12$ you can find the value of $\cos\varphi$ by solving the quadratic equation and taking the positive root. Once you know $\cos\varphi$, you can compute $\cos2\varphi$ and many other things. If you try it this way, you can check your result e.g. here.)

Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$

Now, \begin{align*} \cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1 \end{align*} This is a quadratic equation of the form $2 x^{2} - x -1 =0$ and solving this will give you the value of $\cos\frac{\pi}{5}$ from which you can find the above value which you need.

For the first equality.
\begin{split} \cos(\pi/5) - \cos(2\pi/5) &=\cos(3\pi/10-\pi/10) - \cos(3\pi/10+\pi/10)\\ &=2\sin(\pi/10)\sin(3\pi/10)=2\sin(\pi/10)\cos(\pi/5)\\ &=\frac{2\sin(\pi/10)\cos(\pi/10)\cos(\pi/5)}{\cos(\pi/10)}\\ &=\frac{\sin(\pi/5)\cos(\pi/5)}{\cos(\pi/10)}\\ &=\frac{\sin(2\pi/5)}{2\cos(\pi/10)}=\frac 12. \end{split}

Here 's the latter equality: \begin{split} \cos(\pi/5)\cos(2\pi/5) &= \frac{\sin(\pi/5)\cos(\pi/5)\cos(2\pi/5)}{\sin(\pi/5)}\\ &=\frac{\sin(2\pi/5)\cos(2\pi/5)}{2\sin(\pi/5)}\\ &=\frac{ \sin(4\pi/5)}{4\sin(\pi/5)}\\ &=\frac 14.\end{split}