Explanation

(a) KE of a body undergoing SHM is given by
$KE = \frac{1}{2}m\omega^{2}A^{2}\cos^{2}{\omega t}\ and\ KE_{\max} = \frac{m\omega^{2}A^{2}}{2}$
[symbols represent standard quantities]
From given information
KE=$(KE_{\max}) \times \frac{75}{100}$
$\Rightarrow$ $\frac{m\omega^{2}A^{2}}{2}\ \cos^{2}{\omega t} = \frac{m\omega^{2}A^{2}}{2} \times \frac{3}{4}$
$\Rightarrow$ $\cos{\omega t = \pm \frac{\sqrt{3}}{2}}$
$\Rightarrow$ $\omega t = \frac{\pi}{6}$
$\Rightarrow$ $\frac{2\pi}{T} \times t = \frac{\pi}{6}$
$\Rightarrow$ $t = \frac{T}{12} = \frac{1}{6}$ s