The extreme values of $4\cos \left(x^2\right)\cos (\frac{\pi }{3}+x^2)+{\cos }^2(\frac{\pi }{3}-x^2)$  over  are

$\text{Let\hspace{0.17em}\hspace{0.17em}}{x}^2=\theta$

$\begin{array}{l}=4\cos \theta .\cos (60°+\theta )+{\cos }^2(60°-\theta )\\ =2\left(2\cos \theta \right).\cos (60°+\theta )+\frac{1+\cos 2(60°-\theta )}{2}\\ =2\left[\cos (60°+\theta +\theta )\right]+\left[\cos (60°+\theta -\theta )\right]+\frac{1}{2}+\frac{1}{2}\cos (120°-2\theta )\\ =2(\cos (60°+2\theta )+\cos 60°)+\frac{1}{2}+\frac{1}{2}\cos (180°-(60°+2\theta ))\end{array}$

$\begin{array}{l}=\frac{3}{2}\cos (60°+2\theta )+\frac{3}{2}\\ -1\le \cos (60°+2\theta )\le 1\\ \frac{-3}{2}\le \frac{3}{2}\cos (60°+2\theta )\le \frac{3}{2}\\ 0\le \frac{3}{2}+\frac{3}{2}\cos (60°+2\theta )\le 3\end{array}$

$\therefore$Extreme values are 0,3