The Venn diagram below shows the number of learners in a class who play netball, tennis, or volleyball. Given that there are 160 learners in the class.

*a) (i) The completed Venn diagram is as follows:
Let N be Netball, T be Tennis, and V be Volleyball.
The regions are filled as:

• Netball only: 40
• Tennis only: 25
• Volleyball only: 28
• Netball and Tennis only (not Volleyball): $50 - x$
• Netball and Volleyball only (not Tennis): $35 - x$
• Tennis and Volleyball only (not Netball): $30 - x$
• Netball, Tennis, and Volleyball (all three): $x$

(ii) To find the value of $x$, we sum all the disjoint regions in the Venn diagram and equate it to the total number of learners, which is 160.

$$40 + 25 + 28 + (50 - x) + (35 - x) + (30 - x) + x = 160$$

Combine the constant terms:

$$40 + 25 + 28 + 50 + 35 + 30 = 208$$

Combine the terms with $x$:

$$-x - x - x + x = -2x$$

So the equation becomes:

$$208 - 2x = 160$$

Subtract 208 from both sides:

$$-2x = 160 - 208$$ $$-2x = -48$$

Divide by -2:

$$x = \frac{-48}{-2}$$ $$x = 24$$

The value of $x$ is $\boxed{\text{24}}$.

*b) Given the universal set $\xi = \{x : 1 \le x < 10, x \in \mathbb{Z}\}$.
This means $\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
A is a set of odd numbers.
B is a set of factors of 10.

(i) To list all elements of set A:
Set A consists of odd numbers from the universal set $\xi$.

$$A = \{1, 3, 5, 7, 9\}$$

(ii) To find $n(A' \cap B)$:
First, list the elements of set B. Factors of 10 are 1, 2, 5, 10. From $\xi$, B is:

$$B = \{1, 2, 5\}$$

Next, find the complement of A, denoted as A'. A' contains all elements in $\xi$ that are not in A.

$$A' = \{2, 4, 6, 8\}$$

Now, find the intersection of A' and B, denoted as $A' \cap B$. This set contains elements common to both A' and B.

$$A' \cap B = \{2\}$$

Finally, find $n(A' \cap B)$, which is the number of elements in the set $A' \cap B$.

$$n(A' \cap B) = 1$$

The value of $n(A' \cap B)$ is $\boxed{\text{1}}$.