Statement-I: ~(A⇔∼B) is equivalent to A⇔BStatement-2:  A∨(~(A∧~B)) is a tautology.

We have A∨[~(A∧~B)]≡A∨[~A∨B]≡(A∨~A)∨B≡T∨B≡TThus, Statement-2 is true.We have≡~[A⇔~B]=∼[(A→∼B)∧(~B→A)]≡∼[(~A∨~B)∧(B∨A)]≡∼[~(A∧B)∧(A∨B)]≡[(A∧B)∨(~A∧~B)]≡[(A∧B)∨(~A)]∧[(A∧B)∨(~B)]≡[(A∨~A)∨(B∨−A)]∧[(A∨~B)∧(B∨−B)]≡[T∧(~A∨B)]∧[(A∨~B)∧T]≡(A→B)∧(B→A)≡A⇔B