Since, $f(x)=\left\{\begin{array}{cc}m x+1, & \text { if } x \leq \frac{\pi}{2} \\ (\sin x+n), & \text { if } x>\frac{\pi}{2}\end{array}\right.$

is continuous at $x=\frac{\pi}{2}$
$\mathrm{RHL}$
$$
\begin{aligned}
&=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{+}}(\sin x+n)=\lim _{h \rightarrow 0}\left[\sin \left(\frac{\pi}{2}+h\right)+n\right] \\
&=\lim _{h \rightarrow 0}[\cos h+n]=\cos 0+n=1+n \\
&\mathrm{LHL}=\lim _{x \rightarrow\left(\frac{\pi}{2}\right)^{-}}(m x+1)=\lim _{h \rightarrow 0}\left[m\left(\frac{\pi}{2}-h\right)+1\right]=\frac{m \pi}{2}+1
\end{aligned}
$$
As $f(x)$ is continuous at $x=\frac{\pi}{2}$. So $\mathrm{LHL}=\mathrm{RHL}$
$\Rightarrow m \frac{\pi}{2}+1=n+1$ Hence, $n=m \frac{\pi}{2}$