We have,

$\left(2+\sin x\right)\frac{dy}{dx}=-\left(y+1\right)\cos x$

$\Rightarrow \int \frac{dy}{\left(y+1\right)}=-\int \frac{\cos x}{\sin x+2}dx$

Let, $\sin x=t$$\Rightarrow \cos xdx=dt$

$\Rightarrow \log \left|y+1\right|=-\int \frac{dt}{t+2}+C$

where, $C$ is the constant of integration.

$\Rightarrow \log \left|y+1\right|+\log \left|\sin x+2\right|-C=0$

$\Rightarrow \log \left(y+1\right)\left(\sin x+2\right)=C$

Given, $y\left(0\right)=1$$$

$\Rightarrow C=\log 4$.

$\Rightarrow \log \left(y+1\right)\left(\sin x+2\right)=\log 4 .....\left(\mathrm{i}\right)$

Now, put $x=\frac{\mathrm{\pi }}{2}$ in the equation $\left(i\right)$, we get, $\log \left(y+1\right)+\log \left(3\right)-\log \left(4\right)=0$

$\Rightarrow \log \left(y+1\right)=\log \left(\frac{4}{3}\right)$

$\Rightarrow y+1=\frac{4}{3}$

$\Rightarrow y=\frac{4}{3}-1=\frac{1}{3}$.