1. **Find all subsets of C where C = {3, 7}** The subsets of a set with $n$ elements are $2^n$. Since $C$ has 2 elements, it has $2^2 = 4$ subsets. These subsets are: $\emptyset$, ${3}$, ${7}$, ${3,7}$. 2. **Find $A \cap B$ with $A = \{1, 3, 5\}$ and $B = \{3, 5, 7\}$** $A \cap B$ is the set of elements common to both $A$ and $B$. Common elements are $3$ and $5$, so $A \cap B = \{3, 5\}$. 3. **Find $(A - C) \cup B$ where $C = \{3, 7\}$** First, $A - C$ is the set of elements in $A$ but not in $C$. $A = \{1, 3, 5\}$ and $C = \{3, 7\}$ so $A - C = \{1, 5\}$. Now, $(A - C) \cup B = \{1, 5\} \cup \{3, 5, 7\} = \{1, 3, 5, 7\}$. 4. **Given $Y \subset X$, $|X| = 16$, $|Y| = 5$, and $|X'| = 8$, find:** i. $|Y'|$ By definition of complement, $|X'| = 8$ means the universal set $U$ has $|U|$ elements such that: $$|X'| = |U| - |X| = 8\implies |U| = 8 + 16 = 24$$ Since $Y' = U - Y$, $$|Y'| = |U| - |Y| = 24 - 5 = 19$$ ii. Number of elements in the universal set = $|U| = 24$. 5. **Verify propositional logic equivalences via truth tables:** i. $\sim (p \wedge q) = (\sim p) \lor (\sim q)$ (De Morgan's law) ii. $\sim (p \lor q) \lor (\sim p \wedge q) = \sim p$ iii. $\sim (\sim (p \lor q) \lor \sim (p \lor r)) = p \lor (q \wedge r)$ Verification involves constructing truth tables for $p, q, r$ and checking both sides yield the same truth values in all cases. **Final answers:** 1. Subsets of $C$: $\emptyset, \{3\}, \{7\}, \{3,7\}$ 2. $A \cap B = \{3, 5\}$ 3. $(A - C) \cup B = \{1, 3, 5, 7\}$ 4. (i) $|Y'| = 19$ (ii) $|U| = 24$ 5. Verified by truth tables (equivalences hold).