1. Write the differential equation in standard form

Given $f(x)\sin 2x+\sin x-(1+\cos^2 x)f'(x)=0.$ Rearrange: $(1+\cos^2 x)f'(x)-f(x)\sin 2x=\sin x.$ So $f'(x)-\frac{\sin 2x}{1+\cos^2 x}f(x)=\frac{\sin x}{1+\cos^2 x}.$

Using $\sin 2x=2\sin x\cos x$, $f'(x)-\frac{2\sin x\cos x}{1+\cos^2 x}f(x)=\frac{\sin x}{1+\cos^2 x}.$

This is a linear differential equation: $y'+P(x)y=Q(x),$ with $P(x)=-\frac{2\sin x\cos x}{1+\cos^2 x}.$

2. Find the integrating factor

The integrating factor is $\text{IF}=e^{\int P(x)\,dx}=e^{\int -\frac{2\sin x\cos x}{1+\cos^2 x}\,dx}.$

Let $u=1+\cos^2 x \implies du=-2\sin x\cos x\,dx.$ Then $\int -\frac{2\sin x\cos x}{1+\cos^2 x}\,dx=\int \frac{du}{u}=\ln(1+\cos^2 x).$ Hence $\text{IF}=1+\cos^2 x.$

3. Multiply the equation by the integrating factor

Multiplying $f'(x)-\frac{2\sin x\cos x}{1+\cos^2 x}f(x)=\frac{\sin x}{1+\cos^2 x}$ by $1+\cos^2 x$, we get $(1+\cos^2 x)f'(x)-2\sin x\cos x\,f(x)=\sin x.$

The left side is exactly $\frac{d}{dx}\left[(1+\cos^2 x)f(x)\right],$ because $\frac{d}{dx}(1+\cos^2 x)=-2\sin x\cos x.$ So, $\frac{d}{dx}\left[(1+\cos^2 x)f(x)\right]=\sin x.$

4. Integrate

Integrating, $(1+\cos^2 x)f(x)=\int \sin x\,dx=-\cos x+C.$

5. Use the condition $f(0)=0$

At $x=0$, $(1+\cos^2 0)f(0)=-\cos 0+C.$ Since $\cos 0=1$ and $f(0)=0$, $(1+1)\cdot 0=-1+C \implies C=1.$ So $(1+\cos^2 x)f(x)=1-\cos x.$ Therefore, $f(x)=\frac{1-\cos x}{1+\cos^2 x}.$

6. Evaluate at $x=\frac{\pi}{2}$

Since $\cos\frac{\pi}{2}=0,$ we get $f\left(\frac{\pi}{2}\right)=\frac{1-0}{1+0}=1.$

7. Check options

Thus the correct option is: $\boxed{\text{B: }1}.$