$$(A \cup B) \cap \overline{(A \cap B)} = (A \cap \overline{B}) \cup (\overline{A} \cap B)\tag{1}$$
Step (1) $ (A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap \overline{A} \cup \overline{B}$
Yes, correct, but you should keep parentheses: $ (A \cup B) \cap \overline{(A \cap B)} = (A \cup B) \cap (\overline{A} \cup \overline{B})$
Step (2) Not correct: $ (A \cup B) \cap \overline{(A \cap B)}$ = $A \cup (B \cap \overline{A}) \cup \overline{B} $, by associative property.
No: the associative property (like the commutative property) applies to a chain of unions, or a chain of intersections, but not a mixed chains of unions and intersections. This is why parentheses in the first step are needed. You need to use distribution:
$ (A \cup B) \cap \overline{(A \cap B)}$ = $(A \cup B) \cap (\overline{A} \cup \overline{B}) = [(A\cup B) \cap \overline{A}] \cup [(A \cup B) \cap \overline{B}]$
Can you take it from here? You can use distributivity again...then use that fact that $A \cap \overline{A} = \varnothing$. Likewise for $B\cap \overline{B} = \varnothing$. Simplify.
Another approach is to unpack what proving your equality $(1)$ requires:
In general, we prove that for two sets $P, Q$, $$P = Q \iff P \subseteq Q \text{ AND}\;\;Q\subseteq P$$
For your equality $(1)$, that means you can prove the equality by proving:
$$[(A \cup B) \cap \overline{(A \cap B)}] \subseteq [(A \cap \overline{B}) \cup (\overline{A} \cap B)]\tag{2}$$
and by proving $$[(A \cap \overline{B}) \cup (\overline{A} \cap B)] \subseteq [(A \cup B) \cap \overline{(A \cap B)}]\tag{3}$$
Unpack what this means in terms of "chasing elements", for which I'll give you a start:
$(2)$ If $x \in [(A \cup B)\cap \overline{(A \cap B)}]$ then $x \in (A\cup B)$ AND $x \notin (A \cap B)$, which means $(x \in A$ OR $x \in B)$ AND $(x \notin A$ OR $x \notin B)$...
$(3)$ If $x \in [(A\cap \overline{B})\cup (\overline{A} \cap B)]$, then ...$(x \in A \cap \overline{B})$ OR $(x \in \overline{A} \cap B)$ ....
You must first prove 2 cases:
(1) $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
(2) $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Note that in mathematics we use the following symbols:
$\cap=$ AND = $\land$
$\cup=$ OR = $\lor$
Case 1: $A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Let $x \in A \cap (B \cup C) \implies x \in A \land x \in (B \cup C)$
$\implies x \in A \land \{ x \in B \lor x \in C \}$
$\implies \{ x \in A \land x \in B \} \lor\{ x \in A \land x \in C \} $
$\implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies x \in (A \cap B) \cup (A \cap C)$
$\therefore x \in A \cap (B \cup C) \implies x \in (A \cap B) \cup (A \cap C)$
$\therefore A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C)$
Case 2: $(A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
Let $x \in (A \cap B) \cup (A \cap C) \implies x \in (A \cap B) \lor x \in (A \cap C)$
$\implies \{x \in A \land x \in B \} \lor \{ x \in A \land x \in C \}$
$\implies x \in A \land \{ x \in B \lor x \in C\}$
$\implies x \in A \land \{B \cup C \}$
$\implies x \in A \cap (B \cup C)$
$\therefore x \in (A \cap B) \cup (A \cap C) \implies x \in A \cap (B \cup C)$
$\therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C)$
$\therefore A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
Best Answer
draw.io has templates for Venn diagrams. This tool may prove useful. This discussion contains links to several tools one can use to draw Venn diagrams.
Personally, I have used a python library called pyvenn which allows plotting Venn diagrams containing up to $6$ different sets.