Solution to 3.

 

{(x&~y)->x}->y

 

 x  y  ~y  x&~y  (x&~y)->x  {(x&~y)->x}->y
 T  T  F  F  T  T
 T  F  T  T  T  F
 F  T  F  F  T  T
 F  F  T  F  T  F

 

The statement is a contingency, as can be seen by the fact that there are both Ts and Fs in its truth column.