a) If the groups contain respectively $a_1, a_2, \dots, a_{10}$ points, the number of triangles is
$$N = \binom{a_1}{3} + \binom{a_2}{3} + \dots + \binom{a_{10}}{3}.$$
We claim this number is minimal when
$$a_1 = a_2 = \dots = a_{10} = 10,$$
the minimum value being $10\binom{10}{3}$. Indeed, if there exists a group having $m \ge 11$ points, there will exist another group having $n \le 9$ points. Moving a point from the first group into the second, the number of triangles decreases. This last statement follows from the inequality
$$\binom{m}{3} + \binom{n}{3} > \binom{m-1}{3} + \binom{n+1}{3},$$
readily verified by direct computation.

b) Make each group of $10$ points each, subdivided into two subgroups of $5$ points each. Since the positions of the points are irrelevant, we may suppose all these subgroups of $5$ points make up convex pentagons. Colour their sides with colour $c_1$, their diagonals with colour $c_2$, and the segments joining points from different subgroups with colour $c_3$. It is easy to see no monochromatic triangle appears.