As an extension to Donantonio's comment you could do the first part in a process similar to this:
If you have a $n$ points on the circumference of a circle then you could denote these points by the list: $$\{A,B,C,...\}$$
Now to denote a chord you could write it like: $\{A,C\}$ to denote the chord that goes from $A$ to $C$. Now remembering that the number of combinations of length 2 we can select from a list of $n$ (since we don't care about order, $\{A,C\} \equiv \{C,A\}$) is denoted by:
$$ n\choose 2$$
As a hint you could prove this by using induction and with this identity:
$$ {n\choose k}={n-1\choose k-1}+{n-1\choose k} $$
Situation $2$:
For each selection of any $4$ points on the circumference, you can draw the diagram you have. The line segment that joins the adjacent circumference points, could instead join any of the $4$ pairs of adjacent circumference points, so we have $4$ different triangles for each choice of $4$ circumference points.
There are $\binom{n}{4}$ ways to choose the $4$ points, so Situation $2$ contributes:
$\qquad4 \binom{n}{4}$ triangles.
Situation $3$:
For each selection of any $5$ points on the circumference, you can draw the diagram you have. You could choose any of these $5$ points to be a vertex of a Situation $3$ triangle, so we have $5$ different triangles for each choice of $5$ circumference points.
There are $\binom{n}{5}$ ways to choose the $5$ points, so Situation $3$ contributes:
$\qquad5 \binom{n}{5}$ triangles.
Situation $4$:
For each selection of any $6$ points on the circumference, you can draw the diagram you have. There is only one way to construct that internal triangle given these $6$ circumference points.
There are $\binom{n}{6}$ ways to choose the $6$ points, so Situation $4$ contributes:
$\qquad \binom{n}{6}$ triangles.
Best Answer
From OEIS: Number of intersections of diagonals in the interior of regular n-gon.