If $y=y\left(x\right)$ and $\frac{2+\sin x}{y+1}\left(\frac{dy}{dx}\right)=-\cos x,y\left(0\right)=1$, then $y\left(\frac{\pi }{2}\right)$ equals:

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

$2+\sin x=y$

$\cos xdx=dt$

$\Rightarrow \log (y+1)=-\int \frac{dt}{t}$

$\log (y+1)=-\log \left(t\right)+C$

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

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

when $x=0,y=1$

$\Rightarrow \log \left(2\right)=\log \left(\frac{1}{2}\right)+C$

$\Rightarrow C=-2\log \left(\frac{1}{2}\right)=2\log 2$

$\therefore$ when $x=\frac{\pi }{2}$

$\Rightarrow \log (y+1)=\log \left(\frac{1}{2+1}\right)+2\log 2$

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

$\Rightarrow \log (y+1)-\log \left(\frac{4}{3}\right)=0$

$\Rightarrow \log \left(\frac{3}{4}(y+1)\right)=0$

$\Rightarrow$Taking anti log

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

$\Rightarrow 3y+3=4$

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

$\therefore y\left(\frac{\pi }{2}\right)=\frac{1}{3}=\frac{1}{3}$