First problem: yes. B is either married and unmarried.

Second problem: yes. Let:

X = sqrt(2) (irrational)
Y = X^X
Z = Y^X. (Equals 2, so rational)

If Y is rational, then we have X (irr) ^ X (irr) = Y (rat).

If Y is irrational we have Y (irr) ^ X (irr) = Z (rat).

Note: you can show a stronger result with a cardinality argument. Let R be a rational number.

For each irrational X, let Y(X) = log base X (R). The set of Y's is uncountable, so it must contain an irrational.