Help

If it is divisible by 72, we prime factorize 72 to find its distinct factors. They are 2, 3, 4. This means the entire number is divisible by 2, 3, 4

Knowing 541G5072H6 is divisible by them, we consider the divisibility rules.

- Divisible by 2 if the number is even

- Divisible by 3 if the sum of the digits of the number are divisible by 3

- Divisible by 4 if the last two digits of the number are divisible by 4.

This means that the value of H must be 1, 3, 5, 7, or 9 (if you want the entire number to be divisible by 4. Because H is the second to last digit)

So either it is 

541G507216

541G507236

541G507256

541G507276

541G507296

...

Now considering the divisibility rule of 3 (summing up the digits so we can accord to the divisibility rule)

31 + G

33 + G

35 + G

37 + G

39 + G

Now we guess and check the values of G to find what values make the entire number divisible by 3

If H is 7 or 1, the value of G is 2, 5, or 8.

If H is 3 or 9 the value of  G is 0, 3, 6,  9

If H is 5, then the value of G is 4, or 7

Now we must list the distinct pairs of G * H

2, 5, 8, 14, 35, 56, 0, 9, 18, 27, 54, 81, 20

Then find the sum

329 is your answer

Someone please check this it was a long process I might've made a silly error

There are 3 pairs that satisfy the conditions given in the question and are as follows:
G =5   and H =1
G =1   and H =5
G =6    and H=9
So, the distinct product values are:
5 x 1 = 5
6 x 9 = 54
Sum of the 2 distinct products =5 + 54 = 59