1. The first problem describes an NFA with states q0, q1, and q2, where q2 is the accepting state after seeing "aa". 2. The second problem describes an NFA with states q0 to q5, with q4 and q5 as accepting states for "bb" and "bba" respectively. 3. The third problem gives transitions for states q0 to q3 on inputs 'a' and 'b' and asks if these are correct solutions. Since the user asks if these are correct solutions, we analyze the first problem only as per instructions. **Problem 1: Verify the NFA transitions and acceptance condition for the string ending with "aa"** 1. The NFA has states q0 (initial), q1 (seen 'a'), and q2 (seen 'aa'). 2. Transitions: - From q0: on 'a' go to q1, on 'b' stay at q0. - From q1: on 'a' go to q2, on 'b' go to q0. - From q2: on 'a' stay at q2, on 'b' stay at q2. 3. Acceptance: q2 is accepting, meaning the NFA accepts strings that end with "aa". 4. Explanation: - Starting at q0, reading 'a' moves to q1 (seen one 'a'). - Reading another 'a' moves to q2 (seen "aa"). - Once in q2, any further input keeps the NFA in q2, so it accepts any string that has "aa" as a suffix. 5. This matches the intended behavior of accepting strings ending with "aa". **Final answer:** The NFA described in problem 1 correctly accepts strings ending with "aa".