1) Suppose that we have $k$ triangles in some triangulation. By calculating the sum of all angles of these triangles, we have $180^\circ \cdot k$.
The sum of interior angles of $A$ is $180^\circ \cdot 98$.
The sum of angle around each point among 100 points is $360^\circ \cdot 100$. Hence, we have
$$180^\circ \cdot k = 180^\circ \cdot 98 + 360^\circ \cdot 100 \Leftrightarrow k = 298.$$
Each triangle gives 3 edges and among them, there are 100 edges of $A$. Note that the interior edges are double counted, then the number of edges in each triangulation is
$$\frac{3 \cdot 298 - 100}{2} + 100 = 497.$$

2) Fix the polygon $A$, we will prove this problem by induction on $m+n$ of total points, in which $n$ is the number of vertices of $A$ and $m$ is the number of interior points.

For $m+n=3$, which implies that $n=3, m=0$, we just have one triangle with no interior point. We color this triangle by some color and finish this case.

For bigger $m+n$, whenever we add one more point and perform the triangulation, we have two cases:

1. If there are some triangle that share exactly one edge with the polygon $A$, call triangle $X$. Then $X$ is adjacent to at most two other triangles then if these two triangles were colored by different colors, we just need to color $X$ by the third color (in case two triangles were colored by the same color, we color $X$ by a random color among the rest). Since $X$ has one vertex that is the interior point, then we can move this point outside as the vertex of $A$. It is easy to see that polygon $A$ still has $n$ vertices and $m$ decreases by 1. Then we can apply the induction hypothesis.

2. If for all triangles $X$ as described above, the vertices of $X$ are not the interior points then $X$ is adjacent to exactly one triangle. We just color $X$ by the color different from the adjacent one and remove the vertex of $A$ to make number of sides of $A$ decrease by 2 and $n$ decreases by 1. Then we can also apply the induction hypothesis.

Therefore, in all cases, we can change the triangulation of $m+n$ points to triangulation of $m+n-1$ points, which means the problem can be solved by induction completely.

Then we always can draw the triangle by one of three colors that satisfy the given condition. $\square$