The key lies in "knowing the first two games, A and B each won 1 game." ..... so this is a definite event. You don't need to consider the first two games when calculating the probability, or take the probability of the first two games as 1. Also, in a game, A wins and B loses, and their probabilities add up to one, so only one can be considered.

The first question: winning two more games means that the game is one * * * to four. Then, if the winner of the first three games wins the game, then Party A and Party B have already won one game in each of the first two games, and the game can only be ended if both parties win the remaining two games. ...

0.6*0.6+0.4*0.4

The second question: the first situation is to win four games. The first question. In the second type, if you win three of the five games, then A must win the last game and only win any of the third and fourth games ... so it is 0.6*2(0.6*0.4).