so the water was cooled to 50 degrees from 75 degrees
as far as your other two questions I would've guess so but unfortunately I do not have the ability to test that as i don't have the 5 amp and 30 amp unit. I was hoping to get an idea about how the 30 amp unit would perform with some knowledge about the 5 amp unit.
Well, if you look at that link you will see that
Q = c m dT
where we can take Q to represent your 5 amp or 30 amp unit output over 1 hour, c is some 'constant', m can represent the number of cups of water and dT is the change in temperature. We know that when m=4, dT is about -3.89\(\displaystyle ^{\circ}\)C and Q=5 so that
c
5 ~ 5/4/(-3.89) ~ -0.321
where the subscript 5 represents the 'constant' for 5 amp unit.
Part of our problem is whether c is constant if we change from a 5 amp unit to a 30 amp unit. That is, does the 'caloric output' scale proportionally with the amperage.
If it does, then just use the above equation to compute the change in temperature (in degrees Celsius), i.e.
dT = 30 / (128 c)
EDIT: Note that this is a classic equation and assumes, for example 'a well mixed liquid' as well as several other things. Thus you would have to, at a minimum, 'stir you water gently' so that those assumptions would be at least close to true.