Question 970006
I counted triangles in a systematic, but painful way.
If you know or learn of an elegant solution for this question, I would like to know it.
For the pentagon, I counted 5 diagonals and {{{highlight(35)}}} triangles: {{{drawing(300,300,-2.5,2.5,-1,4,green(circle(0,1.376,0.8)),
triangle(-1,0,1,0,-1.618,1.902),triangle(0,3.078,1,0,1.618,1.902),
triangle(0,3.078,-0.382,1.902,-1.618,1.902),triangle(1.618,1.902,-1,0,-0.382,1.902) )}}}
For the hexagon, after drawing all 9 diagonals (3 long ones, between opposite vertices, and 6 shorter ones),
I counted {{{highlight(110)}}} triangles: {{{drawing(300,300,-2.5,2.5,-2.5,2.5,
green(circle(0,0,1.3)),
triangle(-2,0,1,1.732,1,-1.732),triangle(-1,-1.732,-1,1.732,2,0),
triangle(-1,1.732,1,1.732,0,0),triangle(-1,-1.732,1,-1.732,0,0),
triangle(-1,1.732,-2,0,0,0),triangle(-1,-1.732,-2,0,0,0),
triangle(1,1.732,2,0,0,0),triangle(1,-1.732,2,0,0,0)
)}}}
All those triangles include at least one of the vertices of the polygon (Some include 2 or 3), because excluding all the vertices of the polygon restricts you to the inside of the green circles, and there are no triangles in there.
 
Here is my systematic count, in words:

For the pentagon,
1)	Using 3 vertices of the pentagon to form the triangle:
•	3 adjacent vertices form a triangle, and any of the 5 vertices can be chosen as the middle vertex, forming {{{red(5)}}} different triangles that way.
•	Two adjacent vertices, and a non-adjacent vertex:
There cannot be 3 non-adjacent vertices in a pentagon, but we can make a triangle using 2 adjacent vertices of the pentagon, and a third non-adjacent pentagon  vertex. That means that a side of the pentagon will be a base for that triangle, and the one non-adjacent  pentagon vertex will be the third triangle vertex. Using each pentagon side as a base, {{{red(5)}}} different triangles can be made that way.
2)	Using only 2 vertices of the pentagon to form the triangle:
The third triangle vertex must be the intersection of 2 diagonals.
•	If the two vertices are adjacent, a side of the pentagon will be a base for that triangle, and the third vertex has to be in the intersection of diagonals coming from the ends of that base. Since 2 diagonals come from each end, and each intersects all other diagonals, there are 4 intersection points for the two pairs of diagonals, but one is a vertex of the pentagon. Consequently, for each  side of the pentagon there are 4-1= 3 different triangles that can be made that way. Then, for the whole pentagon, there are 5*3= {{{red(15)}}} different triangles that can be made that way.
•	If the two vertices are non-adjacent, the diagonal connecting them will be a base for that triangle, and the third vertex must be the intersection of the other two diagonals coming from either end of the diagonal acting as the base. There is only 1 such triangle per diagonal, and {{{red(5)}}} different triangles can be made that way.
3)	Using only one pentagon vertex as a triangle vertex, neither a side of the pentagon, nor another vertex of the pentagon can be included in the triangle. Then, the two diagonals coming from the pentagon vertex used form 2 sides of the triangle, and the third side is the diagonal that intersects them in the middle of the pentagon. There is only 1 such triangle for each vertex of the pentagon, and {{{red(5)}}} different triangles can be made that way.
Total = {{{5+5+15+5+5=highlight(35)}}}
 
For the hexagon:
1.	Using 3 vertices of the hexagon to form the triangle:
•	3 adjacent vertices form a triangle, and any of the 6 vertices can be chosen as the middle vertex, forming {{{red(6)}}} different triangles that way.
•	3 non-adjacent vertices form a triangle, and two such sets of non-adjacent vertices exist, so {{{red(2)}}} different triangles that can be made that way.
•	2 adjacent vertices, and a non-adjacent vertex can  form a triangle. The two adjacent vertices can be taken as a base for the triange, with 6 options for such base. The other vertex must be one of the 2 ends  of the parallel side of the hexagon. All in all, 6*2= {{{red(12)}}} such triangles that can be made that way.
2.	Using only 2 vertices of the hexagon for the triangle
•	Using 2 adjacent vertices of the hexagon means using a side of the hexagon as a side/base of the triangle. The other 2 sides of the triangle must be segments of the diagonals coming out of the ends of that base. 3 diagonals come out of each end of the base, so taking one diagonal from each end, there are 9 pairs of diagonals that can be made. One such pair is made of parallel lines.  Two other pairs intersect at other vertices of the hexagon, forming triangles including 3 hexagon vertices, which were already counted above. The other 9-3=6 intersections can be chosen as a third vertex for the triangle. For each side of the hexagon used as a base, 6 different triangles can be formed this way, for a total of 6*6= {{{red(36)}}} triangles with a side of the hexagon as a base and a vertex inside the hexagon.
•	Using 2 opposite vertices of the hexagon means using one of the long diagonals as a base for the triangle. The other sides of the triangle would be other diagonals coming out of the ends of the base. From each end, there is one other diagonal to either side of the base. That makes 2 triangles per long diagonal  base, one to each side of the diagonal base. Since there are 3 long diagonals, 3*2= {{{red(6)}}}  triangles can be made this way.
•	 Using 2 hexagon vertices that are neither adjacent nor opposite, means using one of the 6 short diagonals as a base of the triangle. The other two sides of the triangle will be segments of other diagonals coming out of the ends of the base. For each end there are 2 additional diagonals, that intersect at 2*2=4 points. Since 1 of them is another vertex of the hexagon, there are only 4-1= 3 triangles that can be made this way with each short diagonal base. That makes a total of 6*3= {{{red(18)}}} triangles that can be made this way.
3.	Using only 1 vertex of the hexagon for the triangle
•	The other two vertices of the triangle will be intersections of diagonals coming from that vertex with one other diagonal, but they cannot be ends of that diagonal. For the pair of short diagonals coming out of each vertex, there is only 1 other diagonal that will work, forming 1 triangle. For each of the two pairs made of one short and one long diagonal coming from each vertex, there are 2 diagonals that will work, accounting for 2*2=4 triangles. All in all, using only one vertex of the hexagon, we can make 1+4= 5 triangles per hexagon vertex, and 6*5= {{{red(30)}}} for the whole hexagon.
Total = {{{6+2+12+36+6+18+30=highlight(110)}}} .
 
NOTE: This is the kind of problem that the artofproblemsolving website forum would answer with a very short and elegant explanation that maybe we would be able enough to understand.