Given diff. eq is, $\left(\frac{2+\sin x}{1+y}\right) \frac{d y}{d x}=-\cos x$
$$
\begin{aligned}
&\Rightarrow \frac{d y}{1+y}=-\frac{\cos x}{2+\sin x} d x \\
&\Rightarrow \int \frac{1}{1+y} d y=-\int \frac{\cos x}{2+\sin x} d x \\
&\Rightarrow \log (1+y)=-\log (2+\sin x)+\log C
\end{aligned}
$$

$$
\begin{aligned}
&\Rightarrow \log (1+y)+\log (2+\sin x)=\log C \\
&\Rightarrow \log (1+y)(2+\sin x)=\log C \\
&\Rightarrow 1+y=\frac{C}{2+\sin x} \\
&\Rightarrow y=\frac{C}{2+\sin x}-1 \\
&\text { Put } x=0 \text { and } y=1, \\
&1=\frac{C}{2}-1 \Rightarrow C=4
\end{aligned}
$$
On putting $\mathrm{C}=4$ in Eq. (i), we get
$$
\begin{aligned}
&y=\frac{4}{2+\sin x}-1 \quad \therefore y\left(\frac{\pi}{2}\right)=\frac{4}{2+\sin \frac{\pi}{2}}-1 \\
&=\frac{4}{2+1}-1=\frac{1}{3}
\end{aligned}
$$