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

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Second problem: yes.  Let:

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

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

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

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Note: you can show a stronger result with a cardinality argument.
Let R be a rational number.
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For each irrational X, let Y(X) = log base X (R).
The set of Y's is uncountable, so it must contain an irrational.