$5$. Assume that there are $6$ points of the same color. If they form a convex polygon, its angles add up to $720^\circ$, and there are $6$ internal angles formed by three consecutive vertices, or at least one of them is $\geq120^\circ$, contradiction since the triangle would be monochromatic. If they do not form a convex polygon, four points of the same colour exist such that one is inside the triangle formed by the other three. The internal point is the vertex of three angles that add up to $360^\circ$, hence at least one of them is $\geq120^\circ$, contradiction again. However, a regular pentagon does not provide any angle $\geq120^\circ$, or $5$ points of the same colour may coexist without contradiction with the problem statement.

$5$ again, by the same argument, except that this time we take $6$ points with pairwise distinct colors, finding triangles with all three vertices of different colors and one angle $\geq120^\circ$.

constructed as follows: build a very large regular pentagon, and centered at each one of its vertices, draw a very tiny regular pentagon, such that for each vertex of the large pentagon, the vertices of the tiny pentagon have the same color, which is different from vertex to vertex of the large pentagon. The tiny pentagons may be rotated such that no three points are collinear (possible since there is a finite number of points, hence a finite number of possible threesomes of collinear points that can be destroyed by small rotations of the tiny pentagons aroung its centers), and that no angle between points centered at different vertices of the large pentagons can be $\geq120^\circ$ if the large pentagon is large enough and the tiny pentagons are tiny enough (all such angles would at most be $108^\circ+\epsilon$ where $\epsilon>0$ can be decreased arbitrarily by making the large pentagon larger and the tiny pentagons smaller).

$25$