#### Answer

a) $f$ is continuous at $0$ for all $m$.
b) For $m=2$, $f$ is differentiable at $0$.

#### Work Step by Step

a) For $f$ to be continuous at $x=0$, it is necessary that $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$.
We then examine each limit separately.
- For $x\to0^-$, since $x\lt0$, we have $f(x)=\sin2x$.
Therefore, $\lim_{x\to0^-}f(x)=\sin0=0$
- For $x\to0^+$, since $x\gt0$, we have $f(x)=mx$.
Therefore, $\lim_{x\to0^+}f(x)=m\times0=0$
Since $\lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)$, we see that $f(x)$ is already continuous at $x=0$ no matter which value $m$ might take.
b) For $f$ to be differentiable at $x=0$, it is necessary that $\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)$.
We then examine each limit separately.
- For $x\to0^-$, since $x\lt0$, we have $f(x)=\sin2x$
Hence, $f'(x)=\cos2x\times(2x)'=2\cos2x$.
Therefore, $\lim_{x\to0^-}f'(x)=2\cos(2\times0)=2\cos0=2\times1=2$
- For $x\to0^+$, since $x\gt0$, we have $f(x)=mx$.
Hence $f'(x)=m$
Therefore, $\lim_{x\to0^+}f'(x)=m$
It follows that for $\lim_{x\to0^-}f'(x)=\lim_{x\to0^+}f'(x)$, $m$ needs to equal $2$. In other words, for $m=2$, $f$ is differentiable at $0$.