a) The first dot is placed in triangles:
EAB, FCD, ADF, BEF, CEF, DBF, ECA, ADE

b) The second dot is placed in triangles:
AEF, BFA, CFA, DAF

c) We start with a basic hexagon.
Then from just 1 vertex we draw the line to the other vertices. This forms 4 triangles.
If we keep this diagram and add the lines from the next vertex, we get this figure. We must try to replace the dots to get them in as many triangles as possible. We don't want two dots in one triangle, so we move them into another part of their original triangle. When we add a new set of lines, as long as we do not move the dots out of the previous triangle, we only have to worry about the position of the dots in comparison to the new set of lines.
So once again we add the next set of lines to the next vertex of the hexagon. The way we have previously set the dots out, causes the dots to fall perfectly in the right positions so there is one dot per triangle.
Again we added a new set of lines to the next vertex of the hexagon. We don't have to worry about the previous triangles as long as we keep the dots in their previous triangle. But there are two dots in one triangle and none in another.
So we will move the dot.

We can check by checking from each of the vertices of the hexagon, all of the triangles formed, and see how many dots are in each triangle.

Answer: 4 dots

Note: Please enjoy the simplicity and symmetry of this solution. This was definitely a 'one of a kind' answer!