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

The slope of the line joining $(0,3/2)$ and $(1/2,2)$ is:
$$m=\frac{2-3/2}{1/2-0}=\frac{1/2}{1/2}=1$$
Differentiating the curve equation $y=x+\sin y$:
$$\frac{dy}{dx}=1+\cos y\frac{dy}{dx}$$
$$\frac{dy}{dx}(1-\cos y)=1⟹\frac{dy}{dx}=\frac{1}{1-\cos y}$$
Since the tangent is parallel to the line, their slopes are equal:
$$\frac{1}{1-\cos y}=1⟹1-\cos y=1⟹\cos y=0$$
Since $\cos y=0$, $\sin y$ must be $\pm 1$.The point $(a,b)$ lies on the curve:
$$b=a+\sin b⟹b-a=\sin b$$
Taking the absolute value:
$$|b-a|=|\sin b|$$
Since $\cos b=0$, $|\sin b|=1$.Thus, $|b-a|=1$.