There are 5 points on a straight line R and 8 points on a straight line R 'parallel to R. How many convex quadrangles with vertices in 4 of these 13 points are there?
Making $C_{8}^{2}.C_{5}^{2} = 280$ , but why in this case you can't do $C_{13}^{4} - C_{5}^{4} - C_{8}^{4}$? Seems trivial to me, but I can't come to a conclusion
Your first calculation is correct.
As for your second calculation, we cannot have three or more collinear points in the convex quadrilateral. You accounted for the case where all four points are one line, but you also have to account for the $\binom{8}{1}\binom{5}{3} + \binom{8}{3}\binom{5}{1}$ cases in which three of the four points are collinear. Therefore, you should have $$\binom{13}{4} - \binom{5}{4} - \binom{8}{4} - \binom{8}{1}\binom{5}{3} - \binom{8}{3}\binom{5}{1} = \binom{8}{2}\binom{5}{2}$$