If $f\left(x\right)=\cos \left[e^2\right]x+\cos \left[-e^2\right]x$ where $\left[x\right]$ stands for greatest integer function, then

            $=\cos 7x+\cos \left(-8\right)x$

            $=\cos 7x+\cos 8x$

$f\left(\frac{\pi }{2}\right)=\cos \frac{7\pi }{2}+\cos \frac{8\pi }{2}$

            $=\cos \left(\frac{8\pi -\pi }{2}\right)+\cos 4\pi$

            $=\cos \frac{\pi }{2}+1$ $=0+1=1$

$\therefore f\left(\frac{\pi }{2}\right)=1$