Negation of $x \in A \cup B$. What is the negation of $x \in A \cup B$. Is it $x \in (\lnot A \cap \lnot B)$? Or is it, $x \notin (\lnot A \cap \lnot B)$? Or maybe something else?
Another question: how to distribute nonmembership of $x \notin (\lnot A \cap \lnot B)$?  Is it $x \notin \lnot A \land x \notin \lnot B$?
 A: The negation is 
$\lnot (x\in A \cup B) \iff \lnot [(x \in A) \vee (x \in B)] \iff \lnot (x \in A) \land \lnot(x \in B) \iff x \notin A \land x \notin B$
A: One has:
$$\neg(x\in (A\cup B)) ~{= x\notin (A\cup B) \\ = x\in (A\cup B)^\complement\\ = x\in (A^\complement\cap B^\complement) \\ = x\in A^\complement ~\land~ x\in B^\complement \\ = x\notin A~\land~x\notin B\\ = \lnot(x\in A)~\land~\lnot (x\in B)}$$ 
Similarly:
$$\neg(x\in (A\cap B)) ~{= x\notin (A\cap B) \\ = x\in (A\cap B)^\complement\\ = x\in (A^\complement\cup B^\complement) \\ = x\in A^\complement ~\lor~ x\in B^\complement \\ = x\notin A~\lor~x\notin B\\ = \lnot(x\in A)~\lor~\lnot (x\in B)}$$ 

Note: Although set compementation and logical negation are related concepts, we do not typically use $\lnot A$ to denote the set complement of $A$, rather we usually use $A^\complement$ or similar.  $~A', A^\mathsf c, \lower{0.25ex}\complement A\ldots~$
A: If $x$ is not in $A \cup B$, then $x$ is in neither $A$ nor $B$.  Hence $x$ must be outside both $A$ and $B$.  Thus it is in the complement of both $A$ and $B$.  That means $ x \in A^c \cap B^c$  
