Give an example in which for a set $A \subseteq X$, the two sets $f(X \setminus A)$ and $Y \setminus f(A)$ are incomparable. 
Give an example in which for a set $A ⊆ X$, the two sets $f(X \setminus A)$ and $Y \setminus f(A)$ are incomparable (i.e., neither is a subset of the other).

My example:
Take $X =$ {$1,2$}, $Y =$ {$3$}, and $A =$ {$1$}.
So $f(X \setminus A) = \{3\}$, and $Y \setminus f(A) = \phi$.
Hence neither is a subset of the other. Is my example correct? Any other examples?

Your example does not work, since the empty set is a subset of $\{ 3 \}$, and they are comparable. Moreover you didn't define the function $f$.
Here is an example that works.
$$X=Y = \{ 1,2,3,4\} \qquad  \qquad  \ \ \ A= \{ 2,3\}$$ and $f: X \to Y$ is defined by
$$f(1)=f(2)=1 \qquad \qquad \ f(3)=3 \ \qquad \qquad f(4)=4$$
Then
$Y \setminus f(A)= \{ 2,4 \}$ while $f(X \setminus A) = \{ 1,4\}$. As you can see, these two sets are incomparable under inclusion.
