FOr (a), just see that if $3|n$, then $3|\underbrace{{11\cdots1}}_{n}$ such that $n=3k$. Now, group the $\underbrace{111}_{3}$ into k groups. So, that the $\underbrace{{11\cdots1}}_{n}$ is divisible by $111$, then, we have $111=37*3$ and, we are done.

(b) part can be generalised for any prime $p$ other than 2 or 5. Take $p-1$ ones if $p \neq 3$ and use FLT to prove that such a $k$ divides $p-1$. If $p=3$, take $k=3$.

a) $n\equiv 0 \pmod {3} \Leftrightarrow \underbrace{{11\cdots1}}_{n}=111\cdot 10^{n-3}+111\cdot 10^{n-6}+...+111 \Leftrightarrow \underbrace{{11\cdots1}}_{n}=3*37*\sum_{i=0}^{n/3-1}10^{3i}$ b) just see that $41 \mid 11111$ and $1111,111$ are not divisible by $41$ Then $n \equiv 0 \pmod 5 \implies \underbrace{{11\cdots1}}_{n}=11111\cdot 10^{n-5}+11111\cdot 10^{n-10}+...+11111 \implies \underbrace{{11\cdots1}}_{n}=271*41*\sum_{i=0}^{n/5-1}10^{5i}$