Let's analyze and solve the problem step by step based on the given sets.

1st Row (Sets $A$ and $B$):

• $|A| = 4$, $|B| = 2$, and $|A \cap B| = 0$.

Since $|A \cap B| = 0$, the sets $A$ and $B$ are disjoint, meaning they have no elements in common.

• $|A \cup B|$: The union of disjoint sets is simply the sum of their cardinalities: $|A \cup B| = |A| + |B| = 4 + 2 = 6$

• $|A \setminus B|$: The difference $|A \setminus B|$ refers to the number of elements in $A$ that are not in $B$. Since the sets are disjoint, all elements of $A$ are not in $B$, so: $|A \setminus B| = |A| = 4$

2nd Row (Sets $C$ and $D$):

• $|C| = 8$, $|D| = 4$, and $|C \cap D| = 3$.

• $|C \cup D|$: The formula for the union of two sets is: $|C \cup D| = |C| + |D| - |C \cap D| = 8 + 4 - 3 = 9$

• $|C \setminus D|$: The difference $|C \setminus D|$ refers to the number of elements in $C$ that are not in $D$. This is: $|C \setminus D| = |C| - |C \cap D| = 8 - 3 = 5$

3rd Row (Sets $E$ and $F$):

• $|E| = 13$, $|F| = 20$, and $|E \cup F| = 26$.

We are not given $|E \cap F|$ directly, but we can deduce it from the union formula:

$|E \cup F| = |E| + |F| - |E \cap F|$

Substituting the values:

$26 = 13 + 20 - |E \cap F|$

Solving for $|E \cap F|$:

$|E \cap F| = 33 - 26 = 7$

• $|E \setminus F|$: The difference $|E \setminus F|$ refers to the number of elements in $E$ that are not in $F$. This is: $|E \setminus F| = |E| - |E \cap F| = 13 - 7 = 6$

Final Answers:

1. $|A \cup B| = 6$, $|A \setminus B| = 4$
2. $|C \cup D| = 9$, $|C \setminus D| = 5$
3. $|E \cap F| = 7$, $|E \setminus F| = 6$

Would you like any further details or explanations?

Here are 5 related questions:

1. How would the solution change if $|A \cap B| \neq 0$?
2. Can you express the union and intersection formulas using Venn diagrams?
3. What is the meaning of $|A \setminus B|$ in practical terms?
4. How would you calculate $|A \cup B|$ if more than two sets are involved?
5. How does set theory apply to probability?

Tip: Always use the union formula $|A \cup B| = |A| + |B| - |A \cap B|$ when sets overlap.