Step 1:
LHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\large\frac{k\cos x}{\pi-2x}$
Put $x=\large\frac{\pi}{2}$$-h$
LHL=$\lim\limits_{\large h\to 0}\large\frac{k\cos \big(\Large\frac{\pi}{2}-\normalsize h\big)}{\pi-2\big(\Large\frac{\pi}{2}-\normalsize h\big)}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k\sin h}{\pi-\pi+2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k\sin h}{2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k}{2}\frac{\sin h}{h}$
$\quad\;=\large\frac{k}{2}$
Step 2:
RHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\large\frac{k\cos 2x}{\pi-2x}$
Put $x=\large\frac{\pi}{2}$$+h$
RHL=$\lim\limits_{\large h\to 0}\large\frac{k\cos \big(\Large\frac{\pi}{2}+\normalsize h\big)}{\pi-2\big(\Large\frac{\pi}{2}+\normalsize h\big)}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{-k\sin h}{-2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k}{2}\frac{\sin h}{h}$
$\quad\;=\large\frac{k}{2}$
$f\big(\large\frac{\pi}{2}\big)$$=3\rightarrow $Given
$f$ is continuous if $\large\frac{k}{2}$$=3$ or $k=6$