The first prisoner says the colour of the hat directly in front of him. He should be correct half the time ( I assume). The next prisoner now knows the colour of his hat, so is freed. Repeat the process. Six are freed for sure, and probably half the rest.

100%

You chose the first 2 points to be finitely close to each other and then allowed the 3rd one to be infinitely far away. Seems a little inconsistent.

5 minute and 10 minute guy cross the road in 10 minutes.
5 minute guy goes back on his own in 5 minutes.
The 20 and 25 minute guys cross the road in 25 minutes.
The 10 minute guy crosses back in 10 minutes.
The 5 and 10 minute guys cross together in 10 minutes.
Total time: 1 hour

That is actually the correct solution. Most people let the 5 min/10 min guy run up and down (leading to 65 minutes) and fail to make the two slowest guys walk together,

Well you have to start with two points and yes, they will always have a finite distance between them. You could call that one unit, making your first points be X=0,Y=0 (0,0) and X=1, Y=0 (1,0). If you pick a random point, only if X is between 0 and 1 you will most times have a triangle that is not obtuse while every X>1 and X<0 yields an obtuse one

Why can't you start with the 2 points that are furthest away from each other?

Actually, the answer is probably 50%. There are an equal number of real numbers from -INF to +INF as there are between 0 and 1.

There are three groups of four prisoners. The rearmost prisoner looks at the three in front of him. If the two immediately in front of him have the same colour hat, he calls out the colour of the hat of the third (frontmost) prisoner. Otherwise, he calls out the opposite colour of the third prisoner.

Call the prisoners ABCD. B knows if A called out D's colour correctly. If It is correct, then B knows he has the same colour as C. If not, he knows he has the opposite colour to C.

C also knows if A called out D's colour correctly, so he knows if the colour called out by B matches his own or is the other colour.

D knows that if B and C call out the same colour, then his hat must be the colour called out by A. If they call different colours, then D has the opposite colour to the one called by A.

So BCD know their colours and will be freed. A has a 50% chance (I assume) of being right about his own hat.

The three groups of four prisoners will each have three freed and one with a 50% chance.

pi/2 - 1

Mix them together as the dissipation of heat is related to surface area, so having them together would reduce the total surface area exposed to the air, slowing the change in temperature?

Here we go.

Consider the triangle, in fact, consider the concept of triangle. Scale the plane so that the longest side has length = 1. Let X and Y be the lengths of the other sides. It could be that two or more of the sides have the same length.

0 < X,Y <=1

So each triangle can be identified with a point within the unit square. Actually, since X+Y > 1 for every triangle, the point lies within the upper right triangle of the unit square.

{messed up diagram deleted}

For a triangle to be obtuse, we require that X^2 + Y^2 < 1. This is simply the cosine law. So obtuse triangles lie within the unit circle, and above the diagonal line.

The area of the upper right triangle (= 1/2) represents the set of all triangles in the plane. What portion of them are obtuse?

The area of the quarter circle is pi/4. Subtract the area of the lower left triangle ( 1/2) and we have pi/4 – 1/2. This is the area that corresponds to obtuse triangles in the plane.

The probability that a triangle in the plane is obtuse is simply the ratio of the number of obtuse triangles divided by the number of all triangles in the plane. This is the ratio of the corresponding areas that we have calculated. This is simply pi/2 – 1.

I am famous for missing the blatantly obvious when I get into details, so if I have missed something, please let me know.