a. The smallest number of committees in the school is $6$.
If a student joins at most $2$ committees, that student shares a common committee with at most $2(1004 - 1) = 2008 < 2007$ students, which contradicts the assumption. Therefore, each student joins at least $3$ committees. Thus, there are at least
$$\frac{3 \times 2008}{1004} = 6$$
committees.
The minimum value can be attained. For example, partition all students into $8$ groups $A, B, \dots, H$, each consisting of $251$ students. We form the following $6$ committees:
$$\{A, B, C, D\}, \{A, E, F, G\}, \{A, B, E, H\}, \{B, F, G, H\}, \{C, D, G, H\}, \{C, D, E, F\}.$$
Each committee contains $251 \times 4 = 1004$ students. One can check that every pair of groups belongs to at least one common committee, and so every pair of students joins a common committee.

b. Yes. One can check that every pair of committees in the example provided in part (a) consists of at most $7$ groups only, and so the union consists of at most
$$251 \times 7 = 1757 < 1800$$
students.