Does $\emptyset \setminus A$ make sense? Let be $A$ set. Does $\emptyset \setminus A$ make sense?
 A: Yes, it does, and is made up of those $x \in \emptyset$ such that $x \notin A$, which means that it is $\emptyset$, no matter how paradoxical this may seem at first sight.
A: Yes; it is (as other posters have pointed out) the empty set.
The notation $B\setminus A$ does not require that everything (or anything) in $A$ is actually there in $B$ to remove.
It just removes from $B$ whatever it has in common with $A$, and if $B$ has nothing in common with $A$ (as will be the case if $B$ is already empty), it simply doesn't remove anything and gives you the original $B$ back.
A: Of course it does. It's still the empty set, though.
The statement $x \in \emptyset \setminus A$ means "$x$ is an element of the empty set, but not an element of $A$". There are no such $x$, which means that $\emptyset \setminus A$ has no elements. In other words, it's the same as $\emptyset$.
Another interpretation: If the intersection of two sets $B, C$ is empty (i.e. $B\cap C = \emptyset$), then $B \setminus C = B$ and $C \setminus B = C$. In this case, we clearly have $\emptyset \cap A = \emptyset$, so $\emptyset \setminus A = \emptyset$.
A: Yes:
\begin{equation}
\varnothing \backslash A = \{x\in \varnothing \mid x\notin A\} = \varnothing
\end{equation}
