Regular expression for a finite automaton appears to be wrong
Yes, you are right.

As given FA is DFA. It is accepting string only $\{\epsilon \}$.

And, complement of given  DFA accepting : 
$\implies (1+0+1(0+1))(0+1)^*= (1+0+(10+11))(0+1)^*$

$\implies 1(0+1)^*+0(0+1)^*+(10+11)(0+1)^* = (0+1)^+ =$ { accepting everything other than $\{\epsilon \}$ string over alphabet $\{0,1\}$}.