= Solution
For $m=0$, <continuity> fails. Take $x_k=(\pi/2+2\pi k)^{-1}$ and $\widetilde x_k=(3\pi/2+2\pi k)^{-1}$. Both tend to zero, but the corresponding function values are $1$ and $-1$. Thus there is no <limit>, and the function is not <differentiable> either, since <differentiability> would imply <continuity>.
For $m=1$, the bound $|x\sin(1/x)|\le|x|$ and the <squeeze theorem> give <continuity> at zero. Its difference quotient is $\sin(1/h)$, however, which has no <limit> by the same two sequences. It is therefore not <differentiable> there.
For $m=2$, the bound $|x^2\sin(1/x)|\le x^2$ proves <continuity>. Its difference quotient is $h\sin(1/h)$, whose absolute value is bounded by $|h|$, so the <derivative> exists and equals zero. The complete classification is
$$
\boxed{\begin{array}{c|c|c}m&\text{continuous at }0&\text{differentiable at }0\\\hline0&\text{no}&\text{no}\\1&\text{yes}&\text{no}\\2&\text{yes}&\text{yes, }f_2'(0)=0\end{array}.}
$$
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