3) IMO A.

Here is my approach:

Given: xyz>0
a) Y>0. It implies that x and Z should be both positive or both negative.
If they are both positive, then x and z^3 would be both positive. Hence xy^2z^3 >0.
If they are both negative, then x and z^3 would be both negative. Hence xy^2z^3 > 0
b) x>0, It implies that y and z should be both positive or both negative.
If they are both positive, then y^2 and z^3 would be positive. Hence xy^2z^3 >0, whereas when they are both negative, then y^2 would be positive while z^3 would be negative. Hence this is insufficient.

2) IMO A.

Out of 89, 45th box is going to be the median. The heaviest box of shelf J is the 45th term.

1) IMO B.
Given statement B, 3^x5^y = 1125. Factorizing 1125 gives 9 and 125 that are 3^2 and 5^3 resp.
So x=2.

Thanks.
Saeed