Explanation

(a)
Let $f(x) = 4\cos{(x^{2})}\cos\left( \frac{\pi}{3} + x^{2} \right)\cos\left( \frac{\pi}{3} - x^{2} \right)$
$= 2\cos\left( x^{2} \right)\left\lbrack \cos{\left( \frac{2\pi}{3} \right) + \cos\left( 2x^{2} \right)} \right\rbrack$
$\left\lbrack \because\ \ \ \ 2\cos{A\cos{B = \cos{(A + B) + \cos(A - B)}}} \right\rbrack$
$= 2\cos{(x^{2})}\left\lbrack - \frac{1}{2} + \cos{(2x^{2})} \right\rbrack$
$= - \cos\left( x^{2} \right) + 2\cos{\left( x^{2} \right)\cos\left( 2x^{2} \right)}$
$= - \cos{\left( x^{2} \right) + \cos\left( 3x^{2} \right) + \cos\left( x^{2} \right)}$
$\Rightarrow \ \ \ \ \ \ f(x) = \cos{(3x^{2})}$ ...(i)
$\Rightarrow \ \ \ \ \ f^{'}(x) = - \left\lbrack \sin\left( 3x^{2} \right) \right\rbrack(6x)$
For extremum, put $f^{'}(x) = 0$
$\Rightarrow \ \ \ \ - \sin{\left( 3x^{2} \right)(6x) = 0}$
$\Rightarrow \ \ \ \ x = 0,\ \sqrt{\pi}$
Put $x = 0,\ \sqrt{\pi}$ in Eq. (i), we get
$f(0) = \cos(0) = 1$
And $f(\pi) = \cos{(3\pi)} = - 1$