Step 1: Solve Question 12. Given the sets: $A = \{1, 2\}$ $B = \{2, 3\}$ $C = \{3, 4\}$ a) Compute $A \cup (B \cap C)$ First, calculate the intersection of $B$ and $C$: $$B \cap C = \{2, 3\} \cap \{3, 4\} = \{3\}$$ Next, calculate the union of $A$ with the result: $$A \cup (B \cap C) = \{1, 2\} \cup \{3\}$$ $$A \cup (B \cap C) = \{1, 2, 3\}$$ b) Compute $(A \cup B) \cap (A \cup C)$ First, calculate the union of $A$ and $B$: $$A \cup B = \{1, 2\} \cup \{2, 3\} = \{1, 2, 3\}$$ Next, calculate the union of $A$ and $C$: $$A \cup C = \{1, 2\} \cup \{3, 4\} = \{1, 2, 3, 4\}$$ Finally, calculate the intersection of these two results: $$(A \cup B) \cap (A \cup C) = \{1, 2, 3\} \cap \{1, 2, 3, 4\}$$ $$(A \cup B) \cap (A \cup C) = \{1, 2, 3\}$$ The answers for Question 12 are: *a) $\boxed{\{1, 2, 3\}}$ *b) $\boxed{\{1, 2, 3\}}$ Step 2: Solve Question 13. Given the sets: $A = \{1, 2\}$ $B = \{a, b\}$ Find the Cartesian product $A \times B$. The Cartesian product $A \times B$ is the set of all ordered pairs $(x, y)$ where $x \in A$ and $y \in B$. $$A \times B = \{(1, a), (1, b), (2, a), (2, b)\}$$ The answer for Question 13 is: $\boxed{\{(1, a), (1, b), (2, a), (2, b)\}}$ Step 3: Solve Question 14. Given the sets: $A = \{1, 2, 3\}$ $B = \{a, b, c\}$ A function $f: A \to B$ is defined such that $f(1) = a$, $f(2) = b$, and $f(3) = c$. Determine if this function is injective. A function is injective (one-to-one) if every distinct element in the domain maps to a distinct element in the codomain. In other words, if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. In this case: $f(1) = a$ $f(2) = b$ $f(3) = c$ Each element in $A$ (1, 2, 3) maps to a unique element in $B$ (a, b, c). No two distinct elements in $A$ map to the same element in $B$. Therefore, the function is injective. The answer for Question 14 is: $\boxed{\text{Yes, the function is injective.}}$ Step 4: Solve Question 15. Given the sets: $A = \{1, 2\}$ $B = \{2, 3\}$ Show that $A \cup B = B \cup A$. First, calculate $A \cup B$: $$A \cup B = \{1, 2\} \cup \{2, 3\}$$ $$A \cup B = \{1, 2, 3\}$$ Next, calculate $B \cup A$: $$B \cup A = \{2, 3\} \cup \{1, 2\}$$ $$B \cup A = \{1, 2, 3\}$$ Since both $A \cup B$ and $B \cup A$ result in the set $\{1, 2, 3\}$, we have shown that $A \cup B = B \cup A$. The answer for Question 15 is: $\boxed{A \cup B = \{1, 2, 3\} \text{ and } B \cup A = \{1, 2, 3\}, \text{ therefore } A \cup B = B \cup A.}$