(i)

\begin{align} A \cap B & = \{ 2, 4 \} \end{align}

(ii)

(i)

\begin{align} C \cap D & = \{ \text{blue, yellow, pink} \} \end{align}

(ii)

(i)

\begin{align} E \cap F & = \emptyset \end{align}

(ii)

(i)

(ii)

\begin{align} G \cup H & = \{ \text{apple, orange, banana, grape, durian, pear, strawberry} \} \end{align}

(i)

(ii)

\begin{align} I \cup J & = \{ \text{v, w, x, y, z} \} \end{align}

(i)

(ii)

\begin{align} K \cup L & = \{ 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 \} \end{align}

(i)

\begin{align} M & = \{ 1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2 \} \\ & = \{ 1, 4, 9, 16, 25, 36, 49, 64 \} \\ \\ N & = \{ 1^3, 2^3, 3^3, 4^3 \} \\ & = \{ 1, 8, 27, 64 \} \end{align}

(ii)

\begin{align} M \cap N & = \{ 1, 64 \} \end{align}

(iii)

(i)

\begin{align} P & = \{ 8, 16, 24, 32 \} \\ \\ Q & = \{ 4, 8, 12, 16, 20, 24, 28, 32 \} \end{align}

(ii)

\begin{align} P \cap Q & = \{ 8, 16, 24, 32 \} \end{align}

(iii)

(iv)

\begin{align} \text{Yes, since } P \text{ is a proper subset of } Q \end{align}

(i)

\begin{align} R & = \{ 1, 2, 3, 6, 9, 18 \} \\ \\ S & = \{ 10, 12, 14, 15, 16 \} \phantom{000000} [\text{Composite numbers have more than 2 factors}] \end{align}

(ii)

\begin{align} R \cap S & = \emptyset \end{align}

(iii)

(i)

\begin{align} T & = \{ 3, 6, 9, 12, 15, 18 \} \\ \\ V & = \{ 1, 2, 3, 6, 9, 18 \} \end{align}

(ii)

(iii)

\begin{align} T \cup V & = \{ 1, 2, 3, 6, 9, 12, 15, 18 \} \end{align}

(i)

\begin{align} W & = \{ 4, 8, 12 \} \\ \\ X & = \{ 1, 2, 3, 4, 6, 8, 12, 24 \} \end{align}

(ii)

(iii)

\begin{align} W \cup X & = \{ 1, 2, 3, 4, 6, 8, 12, 24 \} \end{align}

(iv)

\begin{align} \text{Yes, since } W \text{ is a proper subset of } X \end{align}

(i)

\begin{align} Y & = \{ 1, 5, 25 \} \\ \\ Z & = \{ 6, 12, 18, 24 \} \end{align}

(ii)

(iii)

\begin{align} Y \cup Z & = \{ 1, 5, 6, 12, 18, 24, 25 \} \end{align}

\begin{align} y & = x^2 - 3x + 2 \\ \\ \text{Let } & y = 0, \\ 0 & = x^2 - 3x + 2 \\ 0 & = (x - 1)(x - 2) \end{align} \begin{align} x - 1 & = 0 && \text{ or } & x - 2 & =0 \\ x & = 1 &&& x & = 2 \\ \\ \therefore & \phantom{.} (1, 0) &&& \therefore & \phantom{.} (2, 0) \end{align}
\begin{align} \therefore A \cap B & = \{ (1, 0), (2, 0) \} \\ \\ A \cap B & = \{ (x, y) : (x, y) \text{ are the points on the curve where } y = 0 \} \end{align}

\begin{align} y & = x^2 + x + 2 \phantom{00} \text{--- (1)} \\ \\ y & = 3x + 5 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute} & \text{ (2) into (1),} \\ 3x + 5 & = x^2 + x + 2 \\ 0 & = x^2 - 2x - 3 \\ 0 & = (x + 1)(x - 3) \end{align} \begin{align} x + 1 & = 0 && \text{ or } & x - 3 & = 0 \\ x & = -1 &&& x & = 3 \\ \\ \text{Substitute } & \text{into (2),} &&& \text{Substitute } & \text{into (2),} \\ y & = 3(-1) + 5 &&& y & = 3(3) + 5 \\ y & = 2 &&& y & = 14 \\ \\ \therefore & \phantom{.} (-1, 2) &&& \therefore & \phantom{.} (3, 14) \end{align}
\begin{align} C \cap D & = \{ (-1, 2), (3, 14) \} \end{align}

\begin{align} y = (x - 5)^2 + 3 & \rightarrow \text{Minimum curve with turning point } (5, 3) \\ \\ y = (x + 2)^2 - 4 & \rightarrow \text{Minimum curve with turning point } (-2, -4) \\ \\ \\ \therefore E \cup F & = \{ y: y \text{ is a real number such that } y \ge - 4 \} \end{align}