Explanation:

Since HCF = 8 is highest common factor among those numbers
So, let first no = 8x, 2nd number = 8y
So 8x + 8y = 128
x + y = 16
the co-prime numbers which sum up 16 are (1,15), (3,13), (5,11), and (7,9)
*co-prime numbers are those which do no have any factor in common.

These pairs are taken here because in HCF highest common factor is taken,

so the remaining multiples x and y must have no common factors.
Since there are 4 pairs of co-prime numbers here,

there will be 4 pairs of numbers satisfying given conditions.

These are (8*1, 8*15), (8*3, 8*13) , (8*5, 8*11) , (8*7, 8*9)

Hence, option 2 is the correct answer.