(i) and (ii) are simple. (i) If there existed no triangle with three vertices of same colour then the all the points coloured with a colour say $ c_1$ lie of a straight line. We may at max have three distinct lines for each of three colours that do not fill a plane. (ii) We can get a monochromatic triangle with circumradius equal to any positive real number. Pick a arbitrary point in space and construct a circle of radius $ r$, a positive real number. Pick a point on the circumference and let it have the colour $ c_1$ wlog. If we are not able to find another point (or if we find just one more point $ c_1$) on the circumference that is coloured $ c_1$ then every other point on the circumference is coloured either $ c_2$ or $ c_3$. If we do not find a point with $ c_2$ then we are done as rest of the points are $ c_3$. If we find just either 1 or 2 points coloured $ c_2$, then too we have rest of the points coloured $ c_3$.

No no no no... You should prove that there exist a triangle which fullfils all 3 conditions! These are not three diferent problems, these are conditions for triangle

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