This question is a follow-up to this one; see that question for the definition of Banach-Mazur games.

In the absence of dependent choice, determinacy is arguably less natural than quasideterminacy. A quasistrategy lays out a nonempty set of "permitted moves" at each stage, with a quasistrategy s being winning for a player if there is no play consistent with s in which that player loses and a game being quasidetermined if it has a winning quasistrategy.

At the question linked above, Asaf Karagila showed how from a failure of DC we can produce an undetermined Banach-Mazur game. However, the resulting game is still quasidetermined (indeed nicely - player $2$ has a winning quasistrategy and player $1$ does not). My question is whether we can whip up a more extreme counterexample:

Question 1: Is the statement "Every Banach-Mazur game is quasidetermined" consistent with ZF?

Arguably this is actually the wrong question, since quasideterminacy is in one important sense quite weak: it is consistent with ZF that there is a game in which both players have winning quasistrategies! ("Alternate choosing new elements of an amorphous set.") Say that a game is weakly determined if exactly one player has a winning quasistrategy.

Question 2: Is the statement "Every Banach-Mazur game is weakly determined" consistent with ZF?