Marking the first point of breaking the stick with x, and the second with y, there are two cases, x > y and y > x. If 1 > x > y > 0, the stick lengths are y, x-y, and 1-x. "No triangle" happens if one of the sticks is longer than the other two put together, so
1 > x > y > 0 and
y > x-y+1-x
x-y > 1-x+y or
1-x > y+x-y
1 > y > x > 0 and
x > y-x+1-y
y-x > 1-y+x or
1-y > x+y-x
The easiest way I can think of to solve the likelihood is by geometric probability, by finding the matching areas in a 1x1 rectangle where point (x,y) either gives or doesn't give a triangle. It makes a rather pretty picture too.
As for acute or not, it requires the Pythagoras, finding whether c^2 is less than or more than a^2 + b^2.
The rope always forms a triangle, unless two points at exactly one half length of the rope are chosen. Picking the third point suitably, the triangle that is formed can always be made obtuse.
As for the trapezoid, I think it is best tackled as three separate cases - whether the given sides are parallel and opposite, not opposite, and not parallel so the remaining two sides are the parallel ones.