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.
A partial answer: Yes, a straight line on the flattened strip is the shortest path. Think about what happenes to the total path length as you shift any of the intersection points of this path with a triangle edge: the path gets longer.
Be careful, though. Your strip may connect in such a way that a straight line joining the two points will leave the strip. You’ll have to take that possibility into account for a full solution.
Update: I haven’t proven this, but my intuition tells me that if the straight-line path does leave the strip, then adjusting it to get the shortest path is pretty simple. Find the first edge crossing that takes you off the strip and move this intersection to the vertex of that triangle that’s nearest the ultimate goal. Then, take that as your new starting point, draw the straight line from there and iterate.
Best Answer
To get an interior triangle, you need three chords that form the sides of the triangle. Any two of those chords intersect to form a vertex of the interior triangle. Obviously two intersecting chords cannot share an endpoint, so the six points on the circle must each be the endpoint of one of the three chords forming the triangle.
If the 6 points on the circle are labelled A to F in order, then you must connect A to D, B to E, C to F to get the chords that form the sides of the triangle. Any other choice won't work because every chord must intersect the two others (to get the two vertices on that side of the triangle) and therefore has two endpoints on either side. This pairing AD, BE, CF is unique, so there is a unique interior triangle.