Given differential equation is
$$
\frac{(2+\sin x}{(1+y)} \cdot \frac{d y}{d x}=\cos x
$$
which can be rewritten as
$$
\frac{d y}{1+y}=\frac{\cos x}{2+\sin x} d x
$$
Integrate both the sides, we get
$$
\begin{aligned}
& \int \frac{d y}{1+y}=\int \frac{\cos x d x}{2+\sin x} \\
& \Rightarrow \log (1+y)=\log (2+\sin x)+\log C \\
& \Rightarrow 1+y=C(2+\sin x) \\
& \text { Given } y(0)=2 \\
& \Rightarrow 1+2=C[2+\sin 0] \Rightarrow C=\frac{3}{2}
\end{aligned}
$$
Now, $y\left(\frac{\pi}{2}\right)$ can be found as
$$
1+y=\frac{3}{2}\left(2+\sin \frac{\pi}{2}\right) \Rightarrow 1+y=\frac{9}{2}
$$
$$
\Rightarrow \quad y=\frac{7}{2}
$$
Hence, $y\left(\frac{\pi}{2}\right)=\frac{7}{2}$