### Part (a)
1. Using Fermat's Little Theorem (FLT):
- Fermat's Little Theorem states that for any integer $n$ and a prime $p$, $n^p \equiv n \pmod{p}$.
- For $p = 7$, we have $n^7 \equiv n \pmod{7}$. Therefore, $n^{7k} \equiv n \pmod{7}$ for any integer $k$.
- Since $37 = 6 \cdot 6 + 1$, we can write $n^{37} \equiv n^{6 \cdot 6 + 1} \equiv n \pmod{7}$.

2. Using Fermat's Little Theorem for another prime:
- For $p = 13$, we have $n^{13} \equiv n \pmod{13}$. Therefore, $n^{13k} \equiv n \pmod{13}$ for any integer $k$.
- Since $37 = 12 \cdot 3 + 1$, we can write $n^{37} \equiv n^{12 \cdot 3 + 1} \equiv n \pmod{13}$.

3. Combining the results using the Chinese Remainder Theorem:
- We have shown that $n^{37} \equiv n \pmod{7}$ and $n^{37} \equiv n \pmod{13}$.
- Since 7 and 13 are coprime, by the Chinese Remainder Theorem, $n^{37} \equiv n \pmod{91}$.

4. Conclusion for part (a):
- Therefore, $91$ divides $n^{37} - n$ for all integers $n$.

### Part (b)
1. **Finding the largest $k$ that divides $n^{37} - n$ for all integers $n$:**
- We need to find the largest $k$ such that $n^{37} \equiv n \pmod{k}$ for all integers $n$.
- From part (a), we know that $k$ must be a multiple of 91.
- Additionally, we need to consider other primes and their powers that might divide $n^{37} - n$.

2. **Using the Carmichael function $\lambda(k)$:**
- The Carmichael function $\lambda(k)$ gives the smallest positive integer $m$ such that $n^m \equiv 1 \pmod{k}$ for all integers $n$ coprime to $k$.
- For $k = 91$, $\lambda(91) = \text{lcm}(\lambda(7), \lambda(13)) = \text{lcm}(6, 12) = 12$.
- Since $37 \equiv 1 \pmod{12}$, $n^{37} \equiv n \pmod{91}$.

3. Considering other factors:
- We need to check if there are any other factors of $n^{37} - n$ that are not covered by 91.
- The prime factors of $36$ are $2, 3$, and their powers.
- The largest $k$ that divides $n^{37} - n$ for all $n$ must be a multiple of $2^2 \cdot 3^2 \cdot 7 \cdot 13 = 4 \cdot 9 \cdot 91 = 3276$.

4. Conclusion for part (b):
- The largest $k$ that divides $n^{37} - n$ for all integers $n$ is $3276$.

The final answer is $\boxed{3276}$