Problem A is polynomially reducible to problem B…what can we say about A and B?

Your intuition about “relative hardness” is correct, the underlying mathematics is why III. is true. However your justification about I. is a little off (not wrong, but the reasoning is possibly not accurate). It might help to think about reductions in these terms (everything I’ll talk about here will be polynomial time, so I will leave that out just so I don’t have to type it over and over); a reduction (from A to B) is just an algorithm that turns an instance of A into an instance of B, with the special property that if we know/can figure out/can use magic to find out the answer to the instance of B, we know the answer to the original instance of A. This gives us a way of solving problem A if we already know how to solve B. This is why we say things like “B is at least as hard as A”, because any algorithm for B immediately gives an algorithm for A. To flip it around, if we knew B was “easy”, then A must also be “easy” (for some notion of “easy”). Notice that this relationship is not symmetric – knowing how to solve A doesn’t tell us anything about how to solve B (given that all we have is the single reduction). If we had more information, maybe it would, maybe it wouldn’t, but from the reduction alone we get nothing in that direction. So then for I., we know we can turn an A into a B, and that we can solve an A, but this tells us nothing about B. For III., we already know that A is hard (NP-hard!), and the reduction tells us that B is at least as hard as A, so B must also be NP-hard. The formal truth behind this intuition comes from the definition of NP-hardness. Remember that A being NP-hard means every single problem in NP has a reduction to A. We also know that A reduces to B, so how does this show B is NP-hard? We can compose the reductions. For every C in NP, we can reduce every instance of it to an instance of A, then take that instance of A and use our new reduction to turn it into an instance of B (as an exercise you might want to think about why we still know this is all polynomial time etc.). So this gives us a way of turning every problem C in NP into problem B, which is the definition of NP-hardness. Coming back to II., this one again relies on definitions. NP-completeness is NP-hardness, plus the additional property that the problem is in NP. So saying problem A is NP-complete means problem A is NP-hard and A is in NP. We’ve just seen that the reduction to B tells us that B is NP-hard, so we have that part, but just like part I., where A being in P didn’t tell us whether B was in P or not, A being in NP doesn’t tell us whether B is in NP or not. It might be, but the reduction doesn’t give us this information.