It all depends on what hash function you are using for your strings. The worst case for Rabin-Karp would be a case in which every single substring of the text has an equal hash to the pattern, therefore every single substring would be compared and the algorithm is equivalent to brute force.
The first example would be an example of this since the pattern matches in every case, therefore the algorithm would have to individually compare every substring (Assuming you are looking for every match and don't break out on the first one).
The second example would only be an example of a worst case if the hash of "AAB" matches the hash of "AAA," which is unlikely unless you have a really bad hash function. The algorithm would compare the hash of each substring of "AAA" and "AAB" and determine them to be non equal and bypass them until reaching the end where it hits "AAB" and finds and equal hash, comparing the substrings and determining them to be equal.