Polynomial time reducibility
Three of the options use the same trick: if $C1,C2$ are two classes of languages and $C1 \subseteq C2$; then $L1 \in C2$ doesn't imply that $L1 \notin C1$.

The only choice that cannot be made true is 3: if there is a polynomial time reduction from an undecidable language $L_1$ to a decidable language $L_2$ then $L_1$ becomes decidable, too  (just build a Turing Machine that computes the reduction and then solve the problem using the decider for $L_2$) and this is a contraddiction.