If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form $7^{m}+7^{n}$ is divisible by 5 is

The numbers are probably intended to be independently chosen, with the possibility that $m=n$. The remainder of $7^m$ on division by $5$ is equally likely to be one of $2,4,3,1$, since $\varphi(5)$ divides $100$. So is the remainder of $7^n$.

Whatever the remainder of $7^m$ happens to be, there is a unique value of $7^n$ modulo $5$ that gives us sum $0$ modulo $5$. That value has probability $\frac{1}{4}$.

Remark: If we choose a pair of numbers (no replacement), then the answer changes. We can assume that the numbers are chosen in order, $m$ first. Whatever value of $m$ is chosen, there are $25$ choices for $n$ that will give sum $0$ modulo $7$, so the probability is $\frac{25}{99}$.

$7\equiv 2 \bmod 5, 2^1 \equiv 2,2^2 \equiv 4, 2^3\equiv 3 ,2^4\equiv 1$. morever if $m\equiv a\bmod4$ then $2^m\equiv2^a$. (because $2^4\equiv 1 \bmod5)$

so if you want the sum to be divisible by 5 there are 4 ways for this to happen:

$m\equiv1$ and $n\equiv4 \bmod 4$

$m\equiv4$ and $n\equiv1 \bmod 4$

$m\equiv2$ and $n\equiv3 \bmod 4$

$m\equiv3$ and $n\equiv2 \bmod 4$

So if m is any congruence class then the probability n is the corresponding correct class is $\frac{1}{4}$