Let's see F1 :-
(p<->q)^(~p<->q) this is unsatisfiable , since it will be satisfiable only if both are true .
p<->q is true in only two cases TT and FF .
And in both the cases the other term will be False , so it is unsatisfiable.
Now the other one :-
(p+~q)(~p+q)(~p+~q) = (pq+~q~p)(~p+~q) = ~p~q which is satisfiable.