#39

Consider the number 3 0 2 5. It satisfies several interesting properties:
It has four digits
All four digits are different

If we take the number that is the first two digits and sum it with the number that composes the last two digits, its square is the original number: (30+25)2=3025.
The original riddle asked you to find another such number. There is exactly one other, so altogether two. If the requirement that the digits be different is omitted, a third possibility presents itself. Finally, two more solutions enter into the game if we allow the four digits to include leading zeros, so all in all, five solutions to the relaxed version of this riddle.
It is this relaxed riddle that we want to consider. The riddle this month is composed of two questions. Answer both to be considered a solver.

Question 1:
If the relaxed version of the riddle is asked in an arbitrary base, b, instead of in base 10, how many solutions is it going to have, as a function of b?
Question 2:
Consider the strict (original) version of the riddle, but, once again, asked in base b instead of in base 10. Will there be an answer for every b≥4? Prove your claim.

I am still to solve this