Assuming you're using $A^\prime$ to denote the compliment of $A$, note the following.
Simply use the distributivity laws to note:
$A \cap (B \cup A^\prime) \cap B^\prime = ((A \cap B) \cup (A \cap A^\prime)) \cap B^\prime.$
Now, we know $A\cap A^\prime = \emptyset$, so:
$A \cap (B \cup A^\prime) \cap B^\prime = ((A \cap B) \cup \emptyset) \cap B^\prime.$
Note that $(A \cap B) \cup \emptyset$ is the set containing all of the elements in $A \cap B$ and all of the elements in $\emptyset$. However, there are no elements in $\emptyset$, so $(A \cap B) \cup \emptyset$ simply contains all of the elements in $A$. Therefore, $(A \cap B) \cup \emptyset = A \cap B$. Using this, we have:
$A \cap (B \cup A^\prime) \cap B^\prime = (A \cap B) \cap B^\prime.$
Now, set intersection is distributive, so we see
$A \cap (B \cup A^\prime) \cap B^\prime = A \cap (B \cap B^\prime).$
Similar to before, we know $B \cap B^\prime = \emptyset$, so
$A \cap (B \cup A^\prime) \cap B^\prime = A \cap \emptyset = \emptyset.$
You can use truth tables. Let $0$ indicate $x\not\in X$ and $1$ indicate $x\in X$, where $X$ is a set.
For example, the first distributive law:
$$\begin{array} {ccc|c|c|c|c|c}
A&B&C&(B\cup C)&A\cap (B\cup C)&(A\cap B)&(A\cap C)&(A\cap B)\cup (A\cap C)\\
\hline
0&0&0&0&0&0&0&0\\
0&0&1&1&0&0&0&0\\
0&1&0&1&0&0&0&0\\
0&1&1&1&0&0&0&0\\
1&0&0&0&0&0&0&0\\
1&0&1&1&1&0&1&1\\
1&1&0&1&1&1&0&1\\
1&1&1&1&1&1&1&1
\end{array}
$$
Column $5$ equals column $8$.
In response to the updated question, look at the definitions of union and intersection:
$$A\cup B = \{x|x\in A \lor x\in B\}$$
$$A\cap B = \{x|x\in A \land x\in B\}$$
Both $\lor$ and $\land$ are commutative for example (but you need a truth table to prove it), and so, by extending this concept over the definitions of $\cup$ and $\cap$, are these operators, i.e.:
$$A\cup B = \{x|x\in A \lor x\in B\} = \{x|x\in B \lor x\in A\} = B\cup A$$
Best Answer
Hint:
$$(A\cap A')=\emptyset$$
All the steps below (to check your answer or if you give up ...)