Solution:

The sleeping baby is marked in the picture below.

Solution:

one fish, one cat and one dog.

Explanation:
The question might appear a bit difficult in the starting but if you analyze the statements, you will realize that it is just a tricky one.

All except two were fishes and all except two were cats. With these statements we can assume that two of the animals were not fishes and two were not cats. Now one of those animals that are not fishes can be a cat and one of those two animals that are not cats can be a fish. Just carry out the same analysis for the statement that all except two animals were not dogs and you will come across the result i.e.:
In that competition, there was one fish, one cat and one dog.

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:

The answer is 2. First, divide the coins into 3 equal piles. Place a pile on each side of the scale, leaving the remaining pile of 3 coins off the scale. If the scale does not tip, you know that the 6 coins on the scale are legitimate, and the counterfeit is in the pile in front of you. If the scale does tip, you know the counterfeit is in the pile on the side of the scale that raised up. Either way, put the 6 legitimate coins aside. Having only 3 coins left, put a coin on each side of the scale, leaving the third in front of you. The same process of elimination will find the counterfeit coin.

Solution:

A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

Solution:

Can be found on the pic below.

Solution:

Numbers of the first cube:
1, 2, 4, 0, 5 and 6.

Numbers on the second cube:
3, 1, 2, 7, 8 and 0.

Now you can pick any date and you will be able to show it using these cubes. Remember that you can always use 6 as 9 by inverting it.

This problem has another solution and we leave it for you to deduce.

Solution:

Sam : 2
Dean : 5
Castiel : 7

Age diff
7 - 2 = '5' is prime
7 - 5 = '2' is prime
5 - 2 = '3' is prime

Solution:

fIvE thOUsAnd(5000)