The correct answer is: assertion 3.
In all my answer, I will assume that there are 3 voters or more. (With 1 voter, the system is obviously dictatorial. And with 2 voters, the answer to your question depends on the exact tie-breaking rule that is used.)
Remark that a subset consisting of more than half the voters can always choose the winning candidate by giving her the best grade and attributing the worst grade to all the other candidates.
- This precludes the existence of a dictatorial voter, so assertion 1 is false.
- This proves that any candidate can be elected, so assertion 2 is false.
Hence (still with the assumption of having 3 voters or more), Gibbard's theorem implies that as soon as there are 3 candidates or more, MJ is susceptible to tactical voting.
But there is worse. For example, consider the following situation. I use grades for clarity, but the example can be immediately translated with appreciations instead.
Voter 1: A 10, B 0.
Voter 2: B 7, A 0.
Voter 3: B 9, A 8.
Candidate A's median: 8. Candidate B's median: 7. So, A wins. But voter 3 can manipulate toward the following situation.
Voter 1: A 10, B 0.
Voter 2: B 7, A 0.
Voter 3: B 9, A 0.
Candidate A's median: 0. Candidate B's median: 7. So candidate B wins.
Conclusion: MJ is manipulable even when there are only 2 candidates!
I never understood why Balinski and Laraki base a large part of their argumentation in favor of MJ on the issue of manipulability. Although MJ does have some interesting features, immunity to manipulation is certainly not one of them.