1. Answer: 1 (in all options).
Solution. (Option 6-1). Let's represent $k$ as $k=10 a+b$, where $a, b$ are integers and $1 \leqslant b \leqslant 9$. Then $k^{2}+6 k=10\left(10 a^{2}+2 a b+6 a\right)+b^{2}+6 b$. The expression $b^{2}+6 b$ ends in 6 only if $b=2$. But in this case $k^{2}+6 k=$ $100\left(a^{2}+a\right)+16$, which means the second-to-last digit is 1.

In option $6-2$, we need to consider that $k^{2}+2 k-8=(k-2)^{2}+6(k-2)$.