= Solution
If $\alpha>1$, then $f_\alpha$ is an <integrable function>, so the <Riemann-Lebesgue lemma> supplies a continuous representative of its <Fourier transform>. Equivalently, in the mixture for $h_\alpha$ the majorant $t^{(\alpha-3)/2}e^{-t}$ is integrable at zero precisely when $\alpha>1$. In this case <dominated convergence> also gives
$$
h_\alpha(0)=\frac{\sqrt\pi\,\Gamma((\alpha-1)/2)}{\Gamma(\alpha/2)}.
$$
If $0<\alpha\leq1$, <Fatou's lemma> applied along any sequence of nonzero $\xi$ tending to zero gives
$$
\liminf_{\xi\to0,\,\xi\ne0}h_\alpha(\xi)
\geq\frac{\sqrt\pi}{\Gamma(\alpha/2)}\int_0^\infty t^{(\alpha-3)/2}e^{-t}\,dt=+\infty.
$$
The divergence comes from zero, where the exponent is at most $-1$. A continuous representative $H$ would agree with $h_\alpha$ <almost everywhere>. Since both are continuous on each open half-line, they must agree everywhere on those half-lines: a nonzero continuous difference would be nonzero on an interval of positive <Lebesgue measure>. The displayed blow-up then contradicts continuity of $H$ at zero. Changing a value at one point cannot remove this singularity. Consequently
$$
\boxed{\widehat{f_\alpha}\text{ has a continuous representative on }\mathbb R\iff\alpha>1.}
$$
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