1. Given differential equation

$$(2+\sin x)\frac{dy}{dx}+(y+1)\cos x=0$$

with initial condition

$$y(0)=1.$$

2. Rewrite in separable form

Divide both sides by $2+\sin x$:

$$\frac{dy}{dx}+\frac{(y+1)\cos x}{2+\sin x}=0$$

So,

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

Now separate variables:

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

3. Integrate both sides

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

Left side:

$$\int \frac{dy}{y+1}=\ln|y+1|.$$

For the right side, let

$$u=2+\sin x \quad \Rightarrow \quad d\nu=\cos x\,dx.$$

Hence,

$$-\int \frac{\cos x}{2+\sin x}\,dx = -\int \frac{d\nu}{\nu}=-\ln|\nu|=-\ln(2+\sin x).$$

Therefore,

$$\ln|y+1|=-\ln(2+\sin x)+C.$$

4. Simplify the solution

Combine logarithms:

$$\ln|y+1|+\ln(2+\sin x)=C$$ $$\ln\big(|y+1|(2+\sin x)\big)=C.$$

So,

$$(y+1)(2+\sin x)=K$$

for some constant $K$.

Thus,

$$y+1=\frac{K}{2+\sin x}.$$

5. Use the initial condition

Given $y(0)=1$ and $\sin 0=0$:

$$(1+1)(2+0)=K$$ $$2\cdot 2=4.$$

So $K=4$.

Hence,

$$y+1=\frac{4}{2+\sin x}$$

and therefore

$$y=\frac{4}{2+\sin x}-1.$$

6. Find $y\left(\frac{\pi}{2}\right)$

Since $\sin \frac{\pi}{2}=1$,

$$y\left(\frac{\pi}{2}\right)=\frac{4}{2+1}-1=\frac{4}{3}-1=\frac{1}{3}.$$

7. Check options

The correct option is:

$$\boxed{\frac{1}{3}}$$

which is Option D.