Let y=y(x) be the solution of the differential equation,

$\frac{2+\sin x}{y+1} \frac{d y}{d x}=-\cos x, y>0$

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

By integrating both sides :

$\ell \mathrm{n}|\mathrm{y}+1|=-\ell \mathrm{n}|2+\sin \mathrm{x}|+\ell \mathrm{nK}$

$\Rightarrow y+1=\frac{K}{2+\sin x} \quad(y+1>0)$

$\Rightarrow \mathrm{y}(\mathrm{x})=\frac{\mathrm{K}}{2+\sin \mathrm{x}}-1$

Given $\mathrm{y}(0)=1 \Rightarrow \mathrm{K}=4$

So, $y(x)=\frac{4}{2+\sin x}-1$

$a=y(\pi)=1$

$\left.\left.\mathrm{b}=\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\pi}=\frac{-\cos \mathrm{x}}{2+\sin \mathrm{x}}(\mathrm{y}(\mathrm{x})+1)\right]_{\mathrm{x}=\pi}=1$

So, $(a, b)=(1,1)$