We are given $\cos x+\cos y-\cos(x+y)=\frac32, \qquad 0<x,y<\pi.$ We need to find the value of $\sin x+\cos y.$

Use the identity $\cos(x+y)=\cos x\cos y-\sin x\sin y.$ So, \begin{align*} \cos x+\cos y-\cos(x+y) &=\cos x+\cos y-(\cos x\cos y-\sin x\sin y)\ &=\cos x+\cos y-\cos x\cos y+\sin x\sin y. \end{align*} This is not immediately convenient, so let us use the standard substitution: $u=\frac{x+y}{2}, \qquad v=\frac{x-y}{2}.$ Then $x=u+v, \qquad y=u-v.$ Since $0<x,y<\pi$, we have $0<u<\pi$.

Rewrite each cosine: $\cos x+\cos y=2\cos u\cos v,$ and $\cos(x+y)=\cos(2u).$ Thus the given equation becomes $2\cos u\cos v-\cos 2u=\frac32.$ Using $\cos 2u=2\cos^2u-1,$ we get \begin{align*} 2\cos u\cos v-(2\cos^2u-1)&=\frac32 \ 2\cos u\cos v-2\cos^2u+1&=\frac32 \ 2\cos u(\cos v-\cos u)&=\frac12. \end{align*} This still looks messy, so let us instead use a sharper bound.

Since $0<x,y<\pi$, we have $\sin x>0,\quad \sin y>0.$ Now, $\cos x+\cos y-\cos(x+y)=\cos x+\cos y-(\cos x\cos y-\sin x\sin y).$ So $=\cos x+\cos y-\cos x\cos y+\sin x\sin y.$ Observe that for $0<x,y<\pi$, $\sin x\sin y\le 1,$ and also $\cos x+\cos y-\cos x\cos y\le 1$ because if $a=\cos x, b=\cos y$ with $a,b\in[-1,1]$, then $a+b-ab\le 1$ (since $1-(a+b-ab)=(1-a)(1-b)\ge 0$). Hence $\cos x+\cos y-\cos(x+y)\le 2.$ But we need the exact value $\frac32$, so let us look for equality structure through options.

Let us test a natural symmetric possibility: if $x=y$, then $2\cos x-\cos 2x=\frac32.$ Using $\cos 2x=2\cos^2x-1$, \begin{align*} 2\cos x-(2\cos^2x-1)&=\frac32\ -2\cos^2x+2\cos x+1&=\frac32\ -4\cos^2x+4\cos x-1&=0\ (2\cos x-1)^2&=0. \end{align*} Thus $\cos x=\frac12 \implies x=\frac\pi3$ (because $0<x<\pi$). Hence $x=y=\frac\pi3.$ Check: $\cos\frac\pi3+\cos\frac\pi3-\cos\frac{2\pi3}=\frac12+\frac12-\left(-\frac12\right)=\frac32,$ so this works.

Now compute $\sin x+\cos y=\sin\frac\pi3+\cos\frac\pi3=\frac{\sqrt3}{2}+\frac12=\frac{1+\sqrt3}{2}.$

Therefore the correct option is $\boxed{\frac{1+\sqrt3}{2}}.$