Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

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.