2.5. Let $a_n$ be the number the grasshopper is located at after the $n^{th}$ jump, i.e.
$$a_1 = 1, \quad a_n = 1 + k + \dots + k^{n-1}, \quad n \ge 2.$$
We are looking for all numbers $k$ such that $2015 \nmid a_n$ for all $n = 1, \dots, 2015$.
Suppose that $M(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ gives the remainder 1, and since 2015 is divisible by $d$ we have that $2015 \nmid a_n$ for all $n$. Therefore, all positive integers which are not relatively prime to 2015 comply with the terms of the problem.
If $M(k, 2015) = 1$, we observe the remainders of dividing $a_1, \dots, a_{2015}$ by 2015. If one of them is divisible by 2015, such a $k$ is not good. Otherwise, since there are 2014 possible remainders, at least two numbers give the same remainder. Let these numbers be $a_l$ and $a_m$, $m > l$. In this case, their difference is divisible by 2015. On the other hand, we have that
$$a_m - a_l = k^l + \dots + k^{m-1} = k^l (1 + \dots + k^{m-l-1}) = k^l \cdot a_{m-l}.$$
From $2015 \mid k^l \cdot a_{m-l}$ and $M(k, 2015) = 1$, it follows that $2015 \mid a_{m-l}$, which is in contradiction with the assumption that none of the numbers $a_1, \dots, a_{2015}$ is divisible by 2015. Therefore, if $M(k, 2015) = 1$, the grasshopper will jump into a hole.
To conclude, the only numbers which are suitable for the terms of the problem are those which are not relatively prime to 2015.