Is DNF-Equivalence Problem $\mathsf{NP\mbox{-Hard}}$?
A special case of DNF equivalence is DNF tautology: Given a DNF formula $F$, is it satisfied for all assignments? This can be seen by setting $F_1 = F$ and $F_2$ to be a trivial tautology. CNF non-satisfiability is co-NP-complete. Negating the input formula turns a CNF formula into a DNF formula and vice versa and non-satisfiability into tautology. Thus, DNF tautology is co-NP-complete.