Solution:

It is mentioned that 'all but fifteen pencils' were used by him. This phrase means that except fifteen, all his pencils were used.

Thus, he is left with fifteen pencils.

Solution:

The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.

Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.

Solution:

Michelangelo juggles the diamonds across while crossing the bridge.

Solution:

prinka is 10, in a system that awards five for each vowel.

Solution:

60 minutes

Explanation:
Roy will be able to consume all the five tablets in an hour.

Tablet 1 will be taken in 0 min.
Tablet 2 will be taken in 15 min.
Tablet 3 will be taken in 30 min.
Tablet 4 will be taken in 45 min.
Tablet 5 will be taken in 60 min.

Solution:

Playing chess

Game needs two players so Brother-8 is playing chess with Brother-2

Solution:

8/(1−1/5)

Solution:

He should draw any chit of paper and before unfolding it, he must eat and swallow the chit. In this manner, to determine the judgment of the god, they will have to look at the other chit which will have 'Guilty' written over it. Thus everyone will believe that the chit which he ate had innocent written over it and thus they will believe that he is innocent.

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:

He askes the god to beat him half to death.