Answer
Yes, $f(\pi/2)=-\frac{5}{4}$.
Work Step by Step
See graph. It seems the discontinuity is removable. To find the possible function value at $a=\pi/2$, evaluate the limit at $\theta\to\pi/2, \theta\ne\pi/2$:
$\lim_{\theta\to\pi/2}\frac{5cos\theta}{4\theta-2\pi}=\lim_{\theta\to\pi/2}\frac{-5sin(\pi/2-\theta)}{4(\pi/2-\theta)}=-\frac{5}{4}$
Thus, let $f(\pi/2)=-\frac{5}{4}$; the function's continuity is extended at $a=\pi/2$.
