3)

since no three points on the same line can form a triangle there are only two cases that would meet the requirements, ie either one point is on l2 and 2 on l1 or 1 on l1 and two on l2.

For the first case(1 point on l1)

lets say you fix one vertex as A1  now since the other 2 vertices are to be chosen from the remaining 5 points on the line l2 , it can be done in 5C2*3 ways 

similarly for the second case(1 point on l2)

The remaining two points can be chosen in 3C2*5 ways.

So, total ways = 3C2*5+5C2*3=45