When in doubt, Venn Diagrams.
But I'm not going to make one (graphics are difficult). So I'll describe 1:
You have an image with $4$ distinct commponent's
$X = (A\cup B)^c = $ every thing that is in neither $A$ nor $B$. The "outside".
$Y = (A \setminus B) = $ every thing in $A$ but not $B$. The "unique to $A$". part.
$Z = (B \setminus A) = $ everything in $B$ but not in $A$.
$W = A \cap B$. Everything in $B$ and in $A$.
Let's fill these up with elements to make a bland boring system.
Let's put $1$ in $X$. i.e. neither $A$ nor $B$ have $1$.
Let's put $2$ in $Y$. i.e. $A$ has $2$ but $B$ does not.
Let's put $3$ in $Z$. i.e $A$ and $B$ both have $3$.
And put $4$ in $W$. i.e. $B$ has $4$ but $A$ does not.
So $A = \{2,3\}; B= \{3,4\}$ and $U = \{1,2,3,4\}$.
This is a basic boring class of sets.
1) Says: $x \in A^c$. So $x = \{1,4\}$ and $B^c \cap A = \{2\}$ is not empty. This would imply $x \in (A\cup B)^c = \{1\}$.
In this case, that would mean $4$ can not exist and nothing can be in $B \setminus A$. That would mean $B \subset A$.
That's all 1) is describing. 1) = "$B\subset A$ and $B^c \cap A = A \setminus B $ is not empty" or $B \subsetneq A$.
2) Says: $x \in A^c$. So $x = \{1,4\}$. And $x \in B^c =\{1,2\}$ so $x \in A^c \cap B^c = (A\cup B)^c = \{1\}$. If that is the case then $x \in (A\cup B)^c= \{1\}$.
Um... this is always true.
So 1) = "$B\subsetneq A$" and 2) "$A$ and $B$ are sets" (or any obvious always true statement).
So a counter example would be any sets $A$ and $B$ where $B \not \subset A$.
Our example where $A = \{2,3\}$ and $B = \{3,4\}$ is a fine counter example.