Solution:

The only lockers that remain open are perfect squares (1, 4, 9, 16, etc) because they are the only numbers divisible by an odd number of whole numbers; every factor other than the number's square root is paired up with another. Thus, these lockers will be 'changed' an odd number of times, which means they will be left open. All the other numbers are divisible by an even number of factors and will consequently end up closed.

So the number of open lockers is the number of perfect squares less than or equal to one thousand. These numbers are one squared, two squared, three squared, four squared, and so on, up to thirty one squared. (Thirty two squared is greater than one thousand, and therefore out of range.) So the answer is thirty one.

Solution:

If you see the picture, they formed a pentagonal shape and achieved the required.

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

Solution:

First fill the 8 liters jug complete - 4, 8, 0
Fill the 5 liters jug with 8 liters jug - 4, 3, 5
Pour back the beer from 5 liters jug to 12 liters jug - 9, 3, 0
Pour the 3 liters from 8 liters jug to 5 liters jug - 9, 0, 3
Fill the 8 liters jug completely from 12 liters jug - 1, 8, 3
Fill the 5 liters jug from the 8 liters jug - 1, 6, 5
Pour the entire 5 liters jug back in 12 liters jug - 6, 6, 0

You have successfully split the beer into two equal parts.

Solution:

2 people

Say 2 helpers are Helper1 & Helper2

Helper1 food Helper2 food McDuck's Food
End of Day 1 : 1 4 4
Now 'A' will head back

End of Day 2 : N/A 2 4
Now 'B' will head back

Now Mr. McDuck have 4 day food :-)

Solution:

2, 3, 10 and 10
To maintain the difference of eight and keeping in mind the third and fourth hint, the first and third cards can be 2 and 10 or 10 and 2. But keeping in mind the first hint, we know that the first card is 2 and the third is 10. Similarly, if we see the fifth hint, the difference between the second and fourth card should be 7. Thus we know that the second card is 3 and the fourth card is 10.

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

Fifteen minutes can easily be measure using these two hour glasses.

Step 1: He will start both the hourglass.

Step 2: The moment the eleven minute hourglass is empty, he will invert it.

Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.

Solution:

The attached picture shows you how you can do it. If you were able to deduce any other way, feel free to comment.

Solution:

Six.

There are only five colors, so six socks guarantee that two will be the same color.