Solution:

The final digit of a power of $k$ depends only on the final digit of $k$, so there are 10 cases to consider. These are easy to work out. For $k$ ending in 1, the final digits are $1, 1, 1, 1, \ldots$ For $k$ ending in 2 they are $2, 4, 8, 6, 2, 4, 8, 6, \ldots$, et cetera. In fact all 10 possible final digits repeat after 1, 2 or 4 steps, so in every case the final digit is back where it started every 4 steps. Since 2012 is divisible by 4, the last digit of $k^{2012}$ is the same as the last digit of $k^{4}$.

As $k$ varies, the last digits of $k^{4}$ go through a cycle of length 10: $1, 6, 1, 6, 5, 6, 1, 6, 1, 0$.

For part (1), if we list the last digits of the five summands, we have $1, 6, 1, 6, 5$, whose sum has a last digit of $9$.

For part (2), if we list the last digits of the 2012 summands, we will have 201 copies of the sequence $1, 6, 1, 6, 5, 6, 1, 6, 1, 0$, followed by $1$ and $6$. Since $1 + 6 + 1 + 6 + 5 + 6 + 1 + 6 + 1 + 0 = 33$, the last digit of the original sum is the same as the last digit of $201 \cdot 33 + 1 + 6$, which is $0$.