Summation of cosines 
Find the sum $\displaystyle\sum_{n=1}^\infty\left(\frac{2\cos n}{\sqrt{n}}-\frac{1}{\sqrt{n}}\right)\left(\frac{2\cos n}{n}\right)^3\left(\frac{2\cos n}{\sqrt{n}}+\frac{1}{\sqrt{n}}\right)$.

I tried in many ways, like applying Fourier series, GP formula, $\cos^3x$, conversions but nothing worked out.

Doing some algebraic manipulation we get that your sum $S$ is equal to
\begin{align*}
S &= 8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(2\cos(n) -1 \right)\left(2\cos(n) +1 \right)\\
& =8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}\left(4\cos^2(n) -1 \right)\\
& = 32\sum_{n \ge 1}\frac{\cos^5(n)}{n^4} -8\sum_{n \ge 1}\frac{\cos^3(n)}{n^4}
\end{align*}
Now, expanding the cosines in the numerators we get $\cos^3(x) = \frac{3}{4}\cos(x)  +\frac{1}{4}\cos(3x)$ and $\cos^5(x) = \frac{10}{16}\cos(x) + \frac{5}{16}\cos(3x)  + \frac{1}{16}\cos(5x)$. This means that the sum can thus be written as:
\begin{align*}
S &= 2\left[10\sum_{n \ge 1}\frac{\cos(n)}{n^4} + 5\sum_{n \ge 1}\frac{\cos(3n)}{n^4} + \sum_{n \ge 1}\frac{\cos(5n)}{n^4} -3 \sum_{n \ge 1}\frac{\cos(n)}{n^4} -\sum_{n \ge 1}\frac{\cos(3n)}{n^4}\right]\\
& =2\left[7\sum_{n \ge 1}\frac{\cos(n)}{n^4} + 4\sum_{n \ge 1}\frac{\cos(3n)}{n^4} + \sum_{n \ge 1}\frac{\cos(5n)}{n^4}  \right]
\end{align*}
But since we know from this answer that
$$
\sum_{n\ge 1}\frac{\cos(n\,x)}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2}{12}x^2+\frac{\pi}{12}x^3-\frac 1{48}x^4,\quad \text{for}\;x\in(0,2\pi)
$$
Plugging in $x = 1,3,5$ and simplifying gives the final result
\begin{align*}
S =\boxed{\frac{4 \pi^4}{15}- \frac{34 \pi^2}{3} + 40 \pi-\frac{239}{6}} 
\end{align*}
