= Solution
Write each <set difference> as an <intersection> with a <complement of a set>. Applying <De Morgan's laws> and then the <distributive property> gives
$$
\begin{aligned}
A\setminus(B\cup C)&=A\cap\overline{B\cup C}=A\cap(\overline B\cap\overline C)=(A\setminus B)\cap(A\setminus C),\\
A\setminus(B\cap C)&=A\cap\overline{B\cap C}=A\cap(\overline B\cup\overline C)=(A\setminus B)\cup(A\setminus C).
\end{aligned}
$$
The <complements> turn the inner <union> into an <intersection> and vice versa. Thus
$$
\boxed{A\setminus(B\cup C)=(A\setminus B)\cap(A\setminus C),\quad A\setminus(B\cap C)=(A\setminus B)\cup(A\setminus C).}
$$
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