= Solution
Apply the <distributive property> to the <intersection> with $\overline C$:
$$
\begin{aligned}
(A\cap B)\setminus C&=(A\cap B)\cap\overline C=(A\cap\overline C)\cap(B\cap\overline C),\\
(A\cup B)\setminus C&=(A\cup B)\cap\overline C=(A\cap\overline C)\cup(B\cap\overline C).
\end{aligned}
$$
In the first line the repeated factor $\overline C$ changes nothing. Rewriting in terms of <set differences> yields
$$
\boxed{(A\cap B)\setminus C=(A\setminus C)\cap(B\setminus C),\quad(A\cup B)\setminus C=(A\setminus C)\cup(B\setminus C).}
$$
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