 Solution

Both functions are continuous at every $x>0$. At zero, $f_1(1/(\pi /2+2\pi n))=1$ while $f_1(1/(3\pi /2+2\pi n))=-1$, so $f_1$ is discontinuous. Since $|x\sin (1/x)|\le x\to 0$, $f_2$ is continuous at zero and hence everywhere.