Explanation

(a)
$7^{1} = 7,\ 7^{2} = 49,\ 7^{3} = 343,\ 7^{4} = 2401,\ldots.$
Therefore, for $7^{r},\ r \in N$ the no. ends at unit place 7, 9, 3, 1, 7,…. .
$\therefore 7^{m} + 7^{n}$ will be divisible by 5, if it end at 5 or 0.
But it cannot end at 5.
Also it cannot end at 0.
For this $m$ and $n$ should be as follows :
[table] | $m$ | $n$ ; 1 | $4r$ | $4r + 2$ ; 2 | $4r + 1$ | $4r + 3$ ; 3 | $4r + 2$ | $4r$ ; 4 | $4r + 3$ | $4r + 1$ [/table]
For any given value of $m$, there will be 25 values of $n$. Hence, the probability of the required event is$\ \frac{100 \times 25}{100 \times 100} = \frac{1}{4}$.