For a: There are only four possible parities for a point in the Cartesian plane: (odd, odd) (odd, even) (even, odd) (even, even) By the Pigeonhole Principle, there must be at least two points with the same parity in both coordinates. For b: The smallest area I can find is $ 2.5 $, but I'm not sure if this is the smallest possible area. This is in a national olympiad? : : It seems really easy -- and I'm only in eighth grade!

Well, assuming nonconvex pentagons are okay, $(0,0), (1,1), (2,1), (1,2), (0,1)$ has area $3/2$. This is clearly the minimum by Pick's theorem.

(a) The couple of coordinates of any lattice point can have at most four different parity patterns by the pigeonhole principle, the assertion follows. (b) Any triangulation of the pentagon has three triangles with all integer coordinates. Since any lattice triangle has area at least $\frac 1 2$, the minimum possible area for our pentagon is $\frac 3 2$, which is attained for the set of vertices $(0, 0)$, $(0, 1)$, $(1, 0)$, $(1, 1)$, $(2, -1)$.

well ,for (1) there can be 4 types of pts.,if we classify them in terms of the parity of their co ordinates. namely (odd,odd),(even,even) , (odd,even) ,(even ,odd). then applying PHP gives our desired result.

For convex pentagons... http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=403513