7 possible pairs
To solve this, you need to know the divisbility rules of 3 and 5.

Divisibility rule for 3
If the sum of the digits of the number is divisible by 3, then the number is divisible by 3
=> If 357P25Q is divisible by 3
=> 3 + 5 + 7 + P + 2 + 5 + Q is divisible by 3
=> 22 + P + Q is divisible by 3
Divisibility rule for 5
For a number to be divisible by 5, the last digit should be divisible by 5
=> Last digit of the number should be 0 or 5
=> Q = 0 or 5
Now, if Q = 0
=> 22 + P should be divisible by 3
=> P = 2 or 5 or 8
=> There are three possible pairs of {P, Q}

If Q = 5
=> 22 + P + Q should be divisible by 3
=> 27 + P should be divisble by 3
=> P = 0 or 3 or 6 or 9
=> There are four possible pairs of {P, Q}

Total possible pairs of {P, Q} are 3 + 4 = 7
Those pairs are {2, 0} {5, 0} {8, 0} {0, 5} {3, 5} {6, 5} {9, 5}