Given sets H, B, and Q, and student preferences for Mathematics and Physics, calculate unions, intersections, and 'only' preferences.

Step 1: Solve Question 1.
Given the sets:
$H = \{1, 2, 3, 4\}$
$B = \{3, 4, 5, 6\}$
$Q = \{2, 4, 6, 8\}$

a) Find $H \cup B$
The union of two sets includes all elements present in either set.

$$H \cup B = \{1, 2, 3, 4\} \cup \{3, 4, 5, 6\}$$ $$H \cup B = \{1, 2, 3, 4, 5, 6\}$$

b) Find $H \cap Q$
The intersection of two sets includes only the elements common to both sets.

$$H \cap Q = \{1, 2, 3, 4\} \cap \{2, 4, 6, 8\}$$ $$H \cap Q = \{2, 4\}$$

c) Find $(H \cup B) \cap Q$
First, calculate $H \cup B$. From part (a), $H \cup B = \{1, 2, 3, 4, 5, 6\}$.
Next, find the intersection of this result with $Q$.

$$(H \cup B) \cap Q = \{1, 2, 3, 4, 5, 6\} \cap \{2, 4, 6, 8\}$$ $$(H \cup B) \cap Q = \{2, 4, 6\}$$

The answers for Question 1 are:

*a) $\boxed{\{1, 2, 3, 4, 5, 6\}}$

*b) $\boxed{\{2, 4\}}$

*c) $\boxed{\{2, 4, 6\}}$

Step 2: Solve Question 2.
Given:
Total students = 50
Students who like Mathematics ($|M|$) = 30
Students who like Physics ($|P|$) = 20
Students who like both Mathematics and Physics ($|M \cap P|$) = 10

a) How many students like only Mathematics?
To find students who like only Mathematics, subtract those who like both from the total who like Mathematics.

$$\text{Only Mathematics} = |M| - |M \cap P|$$ $$\text{Only Mathematics} = 30 - 10$$ $$\text{Only Mathematics} = 20$$

b) How many students like only Physics?
To find students who like only Physics, subtract those who like both from the total who like Physics.

$$\text{Only Physics} = |P| - |M \cap P|$$ $$\text{Only Physics} = 20 - 10$$ $$\text{Only Physics} = 10$$

The answers for Question 2 are:

*a) $\boxed{20}$

*b) $\boxed{10}$

Step 3: Solve Question 3.
Given the universal set $U$ and sets $A$ and $N$:
$U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ (reordered from the original list)
$A = \{1, 3, 5, 7\}$ (reordered from the original list)
$N = \{2, 4, 6, 8\}$ (reordered from the original list)

a) Find $A^c$ (complement of A)
The complement of A contains all elements in $U$ that are not in $A$.

$$A^c = U - A$$ $$A^c = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 3, 5, 7\}$$ $$A^c = \{2, 4, 6, 8, 9\}$$

b) Find $N^c$ (complement of N)
The complement of N contains all elements in $U$ that are not in $N$.

$$N^c = U - N$$ $$N^c = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{2, 4, 6, 8\}$$ $$N^c = \{1, 3, 5, 7, 9\}$$

c) Find $(A \cup B)^c$
Assuming "B" is a typo and refers to set $N$ from the problem statement, we will calculate $(A \cup N)^c$.
First, find $A \cup N$.

$$A \cup N = \{1, 3, 5, 7\} \cup \{2, 4, 6, 8\}$$ $$A \cup N = \{1, 2, 3, 4, 5, 6, 7, 8\}$$

Next, find the complement of $(A \cup N)$.

$$(A \cup N)^c = U - (A \cup N)$$ $$(A \cup N)^c = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} - \{1, 2, 3, 4, 5, 6, 7, 8\}$$ $$(A \cup N)^c = \{9\}$$

The answers for Question 3 are:

*a) $\boxed{\{2, 4, 6, 8, 9\}}$

*b) $\boxed{\{1, 3, 5, 7, 9\}}$

*c) $\boxed{\{9\}}$ (assuming B refers to N)

Step 4: Solve Question 4.
Given the sets:
$A = \{1, 2, 3, 4, 5\}$
$M = \{3, 4, 5, 6, 7\}$

a) Find $A - M$ (set difference between A and M)
The set difference $A - M$ contains elements that are in A but not in M.

$$A - M = \{1, 2, 3, 4, 5\} - \{3, 4, 5, 6, 7\}$$ $$A - M = \{1, 2\}$$

b) Find $M - A$ (set difference between M and A)
The set difference $M - A$ contains elements that are in M but not in A.

$$M - A = \{3, 4, 5, 6, 7\} - \{1, 2, 3, 4, 5\}$$ $$$$