= Solution
Use the upper half-disc of radius $R$, with a clockwise semicircular indentation of radius $\varepsilon$ above zero. The integrand is <holomorphic> inside that indented contour, so its total integral is zero. On the large arc,
$$
f(z)=\frac{e^{i(a+b)z}-e^{i(a-b)z}}{2iz^2},\qquad |f(z)|\leq R^{-2},
$$
because $a+b>0$, $a-b>0$ and $\operatorname{Im}z\geq0$. Its integral has modulus at most $\pi/R$ and vanishes as $R\to\infty$.
On the small clockwise arc, $f(z)=b/z+O(1)$, so the integral tends to $-i\pi b$. Therefore the remaining real-axis integral satisfies
$$
\operatorname{PV}\int_{-\infty}^{\infty}\frac{e^{iax}\sin(bx)}{x^2}\,dx=i\pi b.
$$
The <imaginary part> is <even> and is an ordinary <absolutely convergent integral>: near zero its integrand tends to $ab$, and at infinity it is $O(x^{-2})$. Taking imaginary parts gives
$$
2\int_0^\infty\frac{\sin(ax)\sin(bx)}{x^2}\,dx=\pi b,\qquad\boxed{\int_0^\infty\frac{\sin(ax)\sin(bx)}{x^2}\,dx=\frac{\pi b}{2}.}
$$
The complex integral needs the <Cauchy principal value> because its <real part> has a $b/x$ <singularity>; the requested sine-product integral does not. This <sine-product integral with inverse-square weight> is an application of the <upper-half-plane indentation rule>.
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