A finite set of coplanar points is such that none lies within the triangle formed by any other three. Are the points vertices of a convex polygon? 
A finite set of points in a plane have a certain property:

If we consider any 3 points in the set, and the triangle generated by these points, then NONE of the other points in the set is in the interior of this triangle.

Does it follow that we can form a convex polygon having all points as vertices?

 A: The answer is No because of "degenerate cases": Take three points forming an equilateral triangle and a fourth point in the interior of one of the edges.
If you exclude such things, using a more careful formulation of the problem, the answer is Yes, because of Carathéodory's theorem. This theorem says that any point in the convex hull of a given set $S\subset{\mathbb R}^2$ is in the convex hull of three points of $S$.
Let $M$ be the set of given points. The convex hull of this set is a convex polygon $P$ with vertex set $S\subset M$, and is then also the convex hull of $S$. When $S$ is not all of $M$ then there is a $p\in M$ which lies in the interior of $P$. This point $p$ would then lie in a triangle formed by points of $S\subset M$, but your clause has  forbidden that.
A: No.
Consider a five-sided polygon: a square with an extra vertex on one side, slightly inwards.  This is a concave polygon, but fulfills the stated criteria: pick any three vertices, and the triangle formed contains none of the other vertices.
Link to illustration
