= Solution
For $a>0$, the fact that <sine> is a <bounded function> gives $|f_a(x)-f_a(0)|\leq |x|^a$. Given $\varepsilon>0$, choose $\delta=\varepsilon^{1/a}$. If $|x|<\delta$, then $|f_a(x)-f_a(0)|<\varepsilon$. This proves <continuity> directly from its definition.
For $a\leq0$, use the first <sequence> from part (b), for which $f_a(x_n)=x_n^a$. When $a=0$ these values are one; when $a<0$ they tend to $+\infty$. In either case, for $\varepsilon=1/2$ every neighbourhood of zero contains a sufficiently late $x_n$ with $|f_a(x_n)-f_a(0)|\geq1/2$. This contradicts the definition of <continuity>. Hence
$$
\boxed{f_a\text{ is continuous at }0\iff a>0}.
$$
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