= Solution
For $F_1$, the functions satisfy $\|\sin(\pi nx)/n\|_\infty\leq1/n$, so the sequence converges uniformly to zero. Its closure is the sequence together with its limit and is compact. The same argument applies to $F_2$, since $\|\sin(\pi nx)/\sqrt n\|_\infty\leq n^{-1/2}$. Thus $F_1$ and $F_2$ are relatively compact.
The family $F_3$ is not <equicontinuous>. At $x=0$ its members vanish, but for $x_n=1/(2n)\to0$,
$$
\sin(\pi n x_n)=1.
$$
By the <Arzelà-Ascoli theorem>, a relatively compact subset of $C[0,1]$ must be equicontinuous. Hence $F_3$ is not relatively compact.
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