= Solution
The <difference quotient> at zero is
$$
\frac{f_a(h)-f_a(0)}h
=\frac{|h|^a}{h}\sin(\pi\sin(1/h)),\qquad h\ne0.
$$
For $a>1$, its <absolute value> is at most $|h|^{a-1}$. Given $\varepsilon>0$, set $\delta=\varepsilon^{1/(a-1)}$. Then $0<|h|<\delta$ makes the <difference quotient> have <absolute value> below $\varepsilon$. Thus $f_a'(0)=0$.
Conversely, along the second <sequence> the <difference quotient> is exactly zero. Any finite <derivative> would consequently have to be zero. Along the first positive <sequence>, however, the <difference quotient> equals $x_n^{a-1}$. It is one for $a=1$ and tends to $+\infty$ for $a<1$. In both cases it stays outside the $\varepsilon=1/2$ neighbourhood of zero for arbitrarily small positive increments. Thus no finite <derivative> exists. We obtain
$$
\boxed{f_a\text{ is differentiable at }0\iff a>1,\qquad f_a'(0)=0\text{ when }a>1}.
$$
In particular, $0<a\leq1$ gives <continuity> without <differentiability>.
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