Let's break down the problem presented in the image and solve it step by step. We are given the following sets:

• $X = \{ 0 \}$
• $A = \{ -1, 0, 1 \}$
• $B = \{ 4, 5 \}$
• $C = \left\{ \frac{p}{q+1} \in \mathbb{Q} \mid p, q \in \mathbb{Z}, 5 \leq p \leq 6 \text{ and } -1 \leq q \leq 1 \right\}$
• $D = \left\{ \frac{4}{5}, \frac{9}{6}, \frac{16}{7}, \frac{25}{8}, \frac{36}{9} \right\}$

Now, let's solve each part.

1. List all elements of $C$

Set $C$ is defined by a rule involving integers $p$ and $q$. The conditions are:

• $p \in \mathbb{Z}$, where $5 \leq p \leq 6$
• $q \in \mathbb{Z}$, where $-1 \leq q \leq 1$

Let's calculate all possible values of $\frac{p}{q+1}$:

For $p = 5$:

• When $q = -1$, $\frac{5}{q+1} = \frac{5}{0}$, which is undefined.
• When $q = 0$, $\frac{5}{q+1} = \frac{5}{1} = 5$.
• When $q = 1$, $\frac{5}{q+1} = \frac{5}{2}$.

For $p = 6$:

• When $q = -1$, $\frac{6}{q+1} = \frac{6}{0}$, which is undefined.
• When $q = 0$, $\frac{6}{q+1} = \frac{6}{1} = 6$.
• When $q = 1$, $\frac{6}{q+1} = \frac{6}{2} = 3$.

Thus, $C = \left\{ 5, \frac{5}{2}, 6, 3 \right\}$.

2. $A \times B$

This represents the Cartesian product of sets $A$ and $B$, meaning we take all possible pairs $(a, b)$ where $a \in A$ and $b \in B$. The pairs are:

$A \times B = \{ (-1, 4), (-1, 5), (0, 4), (0, 5), (1, 4), (1, 5) \}$

3. $A \times A$

Similarly, we calculate the Cartesian product of $A$ with itself:

$A \times A = \{ (-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0), (0, 1), (1, -1), (1, 0), (1, 1) \}$

4. $2^{X \times A}$

This refers to the power set of the Cartesian product $X \times A$. First, we calculate $X \times A$:

$X \times A = \{ (0, -1), (0, 0), (0, 1) \}$

The power set of $X \times A$ is the set of all subsets of $X \times A$. So, the power set is:

$2^{X \times A} = \{ \emptyset, \{(0, -1)\}, \{(0, 0)\}, \{(0, 1)\}, \{(0, -1), (0, 0)\}, \{(0, -1), (0, 1)\}, \{(0, 0), (0, 1)\}, \{(0, -1), (0, 0), (0, 1)\} \}$

5. Describe $A$ and $D$ using set-builder notation

• $A = \{ x \in \mathbb{Z} \mid -1 \leq x \leq 1 \}$
• $D = \left\{ \frac{n^2}{n+4} \mid n \in \mathbb{Z}, 2 \leq n \leq 6 \right\}$

These notations describe $A$ and $D$ using conditions on $x$ and $n$.

Let me know if you'd like further details or explanations.

Here are five related questions to explore further:

1. What is the cardinality of set $C$?
2. How many elements are in the power set $2^{A \times B}$?
3. Can $C$ be expressed using a different interval for $p$ and $q$?
4. How does the Cartesian product change if $B$ had more elements?
5. What is the complement of set $D$ with respect to the rational numbers?

Tip: When working with Cartesian products, remember that the order of elements matters in pairs like $(a, b)$ and $(b, a)$.