Edit: This is now a full solution.
In the below picture, there must be a triangle $R$ which has an edge lying on the red segment, and a triangle $B$ lying on the blue segment. The two sides of $R$ and $B$ lying these colored segments segment cannot both be longer than $1$, else $R$ and $B$ would overlap. Without loss of generality, say that $R$'s side on the red segment is at most one.
Furthermore, considering the side of $R$ on the red edge as the base of $R$, the height of $R$ is at most $7$. Therefore, the area of each triangle is at most $7/2$. This implies that when the board has a single corner removed, the number of triangles is at least $63/(7/2)=18$. This is indeed achievable; tile this checkerboard with a corner removed with $7\times 1$ rectangles, then cut each rectangle in half diagonally.
Credits to antkam in the comments for coming up with this part of the solution.
For the first problem, first suppose that the height of the triangle is at most $6$. Then an area argument proves that at least $20$ triangles are necessary. If we did have height more then $6$, then the triangle would resemble the second picture. Both of the longer sides of the triangle would have length more than $6$. Now, considering the triangles covering the top edge of the chessboard, they would all have to have their small edge on the top (because the other edges are too long), and would be leaning to the left. It is easy to see that this is impossible, and the triangles would have to jut out the left edge of the board. Therefore, $20$ triangles is optimal.