If the tangent to the curve $y=x+\sin y$ at a point $\left(a,b\right)$ is parallel to the line joining  $\left(0,\frac{3}{2}\right)$ and $\left(\frac{1}{2},2\right)$ , then

Given $y=x+\sin y$

$\therefore \frac{dy}{dx}=1+\cos y\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dx}\left(1-\cos y\right)=1\Rightarrow \frac{dy}{dx}=\frac{1}{1-\cos y}\_\_\_\_\left(i\right)$

Since tangent is parallel to line joining $\left(0,\frac{3}{2}\right)$ and $\left(\frac{1}{2},2\right)$

$\therefore {\left[\frac{dy}{dx}\right]}_{(a,b)}=\frac{2-\frac{3}{2}}{\frac{1}{2}-0}=\frac{\frac{1}{2}}{\frac{1}{2}}=1\_\_\_\_\left(ii\right)$

From (i) and (ii), $\frac{1}{1-\cos b}=1$

$\Rightarrow 1-\cos b=1\Rightarrow \cos b=0\Rightarrow b=\left(2n+1\right)\frac{\pi }{2},n\in Z$

Since point $\left(a,b\right)$ lies on curve $y=x+\sin y.$

$\therefore b=a+\sin b\Rightarrow b-a=\sin b\Rightarrow \left|b-a\right|=1$