= Solution
The <convolution> is $(f*g)(x)=\int_{-\infty}^{\infty}f(y)g(x-y)\,dy$. Under the usual integrability hypotheses, <Fubini's theorem> and $u=x-y$ give
$$
\begin{aligned}
\widetilde{f*g}(k)
&=\iint e^{-ikx}f(y)g(x-y)\,dy\,dx\\
&=\left(\int e^{-iky}f(y)\,dy\right)
\left(\int e^{-iku}g(u)\,du\right)
=\boxed{\widetilde f(k)\widetilde g(k)}.
\end{aligned}
$$
The supplied transform and the <Fourier shift theorem> give
$$
\mathcal F\!\left(\frac{\sin x}{x}\right)
=\begin{cases}\pi,&|k|<1,\\0,&|k|>1.\end{cases}
$$
Since
$$
\frac{\sin x}{x^2}=\frac{\cos x}{x}-\frac d{dx}\left(\frac{\sin x}{x}\right),
$$
the shift and differentiation rules yield
$$
\boxed{
\mathcal F\!\left(\frac{\sin x}{x^2}\right)(k)
=\begin{cases}
-i\pi k,&|k|\leq1,\\
-i\pi\,\operatorname{sgn}k,&|k|\geq1.
\end{cases}}
$$
The coincident boundary values make the endpoint convention immaterial.
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