Solution:

a) Let $n = 3m + r$, where $m$ and $r$ are non-negative integers and $r < 3$.
Denote by $\underbrace{00 \ldots 0}_{p}$ a recording with $p$ zeroes and $\underbrace{abcabc \ldots abc}_{p}$ a recording with $p$ times $abc$. We have:
$$\underbrace{11 \ldots 1}_{n} = \underbrace{11 \ldots 1}_{3m + r} = \underbrace{11 \ldots 1}_{3m} \cdot \underbrace{00 \ldots 0}_{r} + \underbrace{11 \ldots 1}_{r} = 111 \cdot \underbrace{100100 \ldots 100}_{m-1} + \underbrace{11 \ldots 1}_{r}.$$
Since $111 = 37 \cdot 3$, the numbers $\underbrace{11 \ldots 1}_{n}$ and $\underbrace{11 \ldots 1}_{r}$ are equal modulo 37. On the other hand, the numbers 1 and 11 are not divisible by 37. We conclude that $\underbrace{11 \ldots 1}_{n}$ is divisible by 37 if and only if $r = 0$, i.e., if and only if $n$ is divisible by 3.

b) Using the idea from a), we look for a repunit which is divisible by 41. Obviously, 1 and 11 are not divisible by 41, while the residues of 111 and 1111 are 29 and 4, respectively. We have $11111 = 41 \cdot 271$. Since 11111 is a repunit with 5 digits, it follows in the same way as in a) that $\underbrace{11 \ldots 1}_{n}$ is divisible by 41 if and only if $n$ is divisible by 5.