Solution:

Sucre : 1 cupcake
T-Bag : 2 cupcakes
Burrows: 3 cupcakes
Scofield : 5 cupcakes

Explanation :
It's obvious if the prisoner has eaten 5 or more cupcakes than he will not ask this question "did you had more cupcakes than me?" as it will make him the person who had eaten maximum cupcakes.
A) Sucre : Must not have eater 5 cupcakes, therefore he has eaten anywhere between 1-4 cupcakes.

B) T-bag : Since sucre asked him the question T-bag knows sucre have eaten between 1 - 4 cupcakes, and therefore he must have eaten 2 to 4 cupcakes(it can't be 1 else he will reply answer as "No".)

C) Burrows : Replied with the answer as "I don't know and based on logic B) he knows T-bag must have eaten at least 2 and he replied no which implies that he must be eaten 3 or 4.

D) Scofield replied he know all, which indicates he must have eaten 5.
Therefore cupcake division is as below:
Sucre : 1
T-Bag : 2
Burrows: 3
Scofield : 5

Solution:

142857

Explanation:
142857 * 1 = 142857
142857 * 2 = 285714
142857 * 3 = 428571
142857 * 4 = 571428
142857 * 5 = 714285
142857 * 6 = 857142

Note: Such numbers are called cyclic numbers.

Solution:

In the attached picture, you can see that the formation allows 2 castles to be protected by the walls. Congrats!

Solution:

It can be solved as below.

Solution:

We just added four matchsticks as shown in below picture.

Solution:

Fred Odea won.

Just look at the announcement lines. The first sentence in 'Ok now everybody!' If you take out just the first letter of every word, it will form 'ONE'. In the same manner the second sentence 'The winner of…..' will give form 'TWO' and the third sentence 'The Hardest Riddle Ever Event.' Will form THREE.

Thus the final sentence must form FOUR. There is only one candidate who can suffice with the first two letter and he is Fred Odea. The final sentence of the mastermind must be, 'Fred Odea: Ultimate Riddler!'

Solution:

64

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

Solution:

fIvE thOUsAnd(5000)

Solution:

They can be expressed as the sum of the cubes of two different sets of numbers.

10^3 + 9^3 = 1729
and
12^3 + 1^3 = 1729