1. Statement of the problem: Given the universal set $\Omega = \{a,b,c,d,e,f,g\}$ and the sets $A=\{a,b,c,d,e\}$, $B=\{a,c,e,g\}$, $C=\{b,e,f,g\}$, find (i) $A \cup B$, (ii) $B \cap A$, (iii) $A \cap (B \cap C)$, and (iv) $(A \cap A')'$. 2. Definitions and rules used: The union $X \cup Y$ is the set of elements that are in $X$ or in $Y$ or in both. The intersection $X \cap Y$ is the set of elements that are in both $X$ and $Y$. The complement $X'$ means all elements in $\Omega$ that are not in $X$. A set intersected with its complement is the empty set, i.e. $A \cap A' = \varnothing$. The complement of the empty set is the universal set, i.e. $\varnothing' = \Omega$. 3. Work for (i) $A \cup B$: List elements of $A$ and $B$ and combine without repetition. $A = \{a,b,c,d,e\}$ and $B = \{a,c,e,g\}$. Combine elements: $\{a,b,c,d,e\} \cup \{a,c,e,g\} = \{a,b,c,d,e,g\}$. So $A \cup B = \{a,b,c,d,e,g\}$. 4. Work for (ii) $B \cap A$: Intersection picks elements common to both $B$ and $A$. $B = \{a,c,e,g\}$ and $A = \{a,b,c,d,e\}$. Common elements are $a,c,e$. So $B \cap A = \{a,c,e\}$. 5. Work for (iii) $A \cap (B \cap C)$: First compute $B \cap C$, then intersect with $A$. $B = \{a,c,e,g\}$ and $C = \{b,e,f,g\}$. $B \cap C = \{e,g\}$ because $e$ and $g$ are in both $B$ and $C$. Now intersect with $A$: $A = \{a,b,c,d,e\}$. $A \cap \{e,g\} = \{e\}$ because only $e$ is in $A$. So $A \cap (B \cap C) = \{e\}$. 6. Work for (iv) $(A \cap A')'$: Use the complement and intersection rules. Compute $A'$ first: $A' = \Omega \setminus A = \{f,g\}$. Now $A \cap A'$ is the intersection of $\{a,b,c,d,e\}$ with $\{f,g\}$. $A \cap A' = \varnothing$ because no element is both in $A$ and in $A'$. Finally take the complement: $(A \cap A')' = \varnothing' = \Omega$. So $(A \cap A')' = \{a,b,c,d,e,f,g\}$. 7. Final answers: (i) $A \cup B = \{a,b,c,d,e,g\}$. (ii) $B \cap A = \{a,c,e\}$. (iii) $A \cap (B \cap C) = \{e\}$. (iv) $(A \cap A')' = \{a,b,c,d,e,f,g\}$.