hoshi911:
You're right. K4 is not necessary. Sorry. Nevertheless, "All inside liberties count for the group with the eye" is incorrect. It is the other way round.

hoshi911:
There is still a problem with the solution. It says: Semeai Type 3: Eye vs no eye. No Seki possible. All inside liberties count for the group with the eye. Black has 6 liberties. White has 5 liberties. White is dead. (1 move behind). As stated before, some of the above made assertions do not apply. I suggest to do a rephrasing: "Semeai Type 3: Eye vs no eye, but with other groups involved. Standard counting principles do not apply. Black needs 5 moves to live (which also kills White). White needs 7 moves to kill. White is dead. (2 moves behind)." Notice that in order to kill Black, White needs to fill 6 six inside liberties O7-T7 and somewhere in between also K4. This makes 7 moves to kill for White.

hoshi911:
solution says: Semeai Type 3: Eye vs no eye. No Seki possible. All inside liberties count for the group with the eye. Black has 6 liberties. White has 5 liberties. White is dead. (1 move behind). What a strange problem! Although the corresponding category "Eye vs no eye" is obviously correct, I doubt that "All inside liberties count for the group with the eye" applies here, because black's only chance is the sequence P1, K1, F1, A3 and finally A5. Therefore, the statement "White has 5 liberties" seems correct (but these do not result from the inside liberties!) . As a matter of fact, the 5 inside liberties P7-T7 (and additionally K4) here count for Black, the group without eye!

hoshi911:
I think there is a construction problem here. After black A3, black can catch the white stones on the left with black A5. White can do nothing about it. So the capturing race on the mid and right hand side is irrelevant.