a.
It is well known that there are $\binom{n+k-1}{k-1}$ ways to divide $n$ marbles into $k$ boxes. In this case, the answer is $\binom{34}{4}$.

b.
Let $m$ be the number of colors, $x_1, x_2, \dots, x_m$ be the number of boxes containing marble with color $1, 2, \dots, m$ respectively. We now count the number of tuples $(A, B, C)$, where $A, B$ are the boxes having marbles with the same color $C$.

On the one hand, since every two boxes have in common at most $3$ colors, thus the number of pairs is at most $3\binom{5}{2} = 30$.
On the other hand, the number of pairs is $S = \sum_{i=1}^{m} \binom{x_i}{2}$. Since in each box, there are at most one marble in each color, we get that $\sum_{i=1}^{m} x_i = 30$. By the Cauchy-Schwarz inequality, we have
$$S = \frac{1}{2} \left( \sum_{i=1}^{m} x_i^2 - \sum_{i=1}^{m} x_i \right) \geq \frac{1}{2} \left( \frac{30^2}{m} - 30 \right).$$
Hence,
$$\frac{900}{m} - 30 \le 60 \Leftrightarrow m \ge 10.$$

c.
Consider the following table.

| Box | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|-----|---|---|---|---|---|---|---|---|---|----|
| 1 | × | × | × | × | × | × | | | | |
| 2 | × | × | × | | | | × | | × | × |
| 3 | × | | | × | × | | | × | × | × |
| 4 | | × | | × | | × | × | × | | × |
| 5 | | | × | | × | × | × | × | × | |

It is a direct checking that the table satisfies the requirements. ☐