1.
We can just check all values of $n\pmod 6$:
| $n$ | $n(n+1)(n+2)$ |
| 0 | 0 |
| 1 | 6 |
| 2 | 24 |
| 3 | 60 |
| 4 | 120 |
| 5 | 210 |
Clearly, all the values of the second column are divisible by $6$. Thus $6\mid n(n+1)(n+2)$.
2.
Note that \begin{align*} & 1^{2015} + 2^{2015} + 3^{2015} + 4^{2015} + 5^{2015} + 6^{2015} \\ \equiv \ & 1^{2015}+2^{2015}+3^{2015} + (-3)^{2015}+(-2)^{2015}+(-1)^{2015} \\ \equiv \ & 0\pmod 7\end{align*} Thus we are done.