a) If $m = 1$, then the class will have 8 group to attend, this is contradiction. If $m \ge 3$, consider any 3 groups $X_1, X_2, X_3$ then
$$n \ge |X_1 \cup X_2 \cup X_3| \ge |X_1| + |X_2| + |X_3| - |X_1 \cap X_2| - |X_2 \cap X_3| - |X_1 \cap X_3| \ge 4 + 4 + 4 - 1 - 1 - 1 = 9, \text{ (contradiction).}$$
So $m = 2$, we can take a simple case that each group has 4 participating classes and no class joins 2 groups.

b) Denote by $S$ the number of sets $(A, \{B, C\})$ in which class $A$ has students participating in groups $B$ and $C$. There are $\binom{6}{2} = 15$ pairs $B, C$ and any two groups have no more than 4 co-participating class so we have $S \le 15 \cdot 4 = 60$.
Let $a_1, a_2, \dots, a_n$ be the number of groups that the student of class 1, 2, ..., $n$ participate to. There are $6 \cdot 10 = 60$ in total of participants so $a_1 + a_2 + \dots + a_n = 60$. We have
$$\begin{aligned} S &= \binom{a_1}{2} + \binom{a_2}{2} + \binom{a_3}{2} + \dots + \binom{a_n}{2} \\ &= \frac{a_1 \cdot (a_1 - 1) + a_2 \cdot (a_2 - 1) + \dots + a_n \cdot (a_n - 1)}{2} \\ &= \frac{1}{2}(a_1^2 + a_2^2 + \dots + a_n^2) - 30. \end{aligned}$$
Since $S \le 60$, we have
$$a_1^2 + a_2^2 + \dots + a_n^2 \le 2(60 + 30) = 180.$$
Applying the Cauchy-Schwarz inequality, we have
$$n(a_1^2 + a_2^2 + \dots + a_n^2) \ge (a_1 + a_2 + \dots + a_n)^2 = 3600.$$
Hence $\frac{3600}{n} \le 180$, which implies that $n \ge 20$.

c) There are a total of $20 \cdot 4 = 80$ participation so there will be a class $D$ with the number of student is at least $\lfloor \frac{80}{n} \rfloor$. And the groups that class $D$ participates in will all have the same 1 co-participant class (class $D$) and the remaining 3 classes of these groups are distinct. Therefore, we get
$$n \ge 3 \left\lfloor \frac{80}{n} \right\rfloor + 1 \ge 3 \cdot \frac{80}{n} + 1 \implies n^2 - n - 240 \ge 0.$$
This implies that $n \ge 16$. We can show a specific case with $n = 16$ as following
$$\begin{aligned} A_1 &= \{1, 2, 3, 4\}, \quad A_2 = \{1, 5, 6, 7\}, \quad A_3 = \{1, 8, 9, 10\}, \\ A_4 &= \{1, 11, 12, 13\}, \quad A_5 = \{1, 14, 15, 16\}, \quad A_6 = \{2, 4, 9, 13\}, \\ A_7 &= \{2, 10, 12, 14\}, \quad A_8 = \{2, 6, 8, 16\}, \quad A_9 = \{2, 7, 11, 15\}, \\ A_{10} &= \{3, 5, 12, 16\}, \quad A_{11} = \{3, 8, 13, 15\}, \quad A_{12} = \{3, 6, 10, 11\}, \\ A_{13} &= \{3, 7, 9, 14\}, \quad A_{14} = \{4, 5, 10, 15\}, \quad A_{15} = \{4, 7, 8, 12\}, \\ A_{16} &= \{4, 9, 11, 16\}, \quad A_{17} = \{4, 6, 13, 14\}, \quad A_{18} = \{5, 8, 11, 14\}, \\ A_{19} &= \{6, 9, 12, 15\}, \quad A_{20} = \{7, 10, 13, 16\}. \end{aligned}$$
So in this case, the minimum value of $n$ is 16. $\square$