The man has 3 daughters; one blonde, one brunette, and one redhead.

It is easy to see that if we have more than 9 balls, we need at least 3 measurements. We will prove that 3 measurements are enough for 12 balls.
We place 4 balls on each side of the scale. Let balls 1, 2, 3, 4 be on the right side, and balls 5, 6, 7, 8 on the left side.
CASE 1. The scale does not tip to any side. For the second measurement we place on the left side balls 1, 2, 3, 9 and on the right side balls 4, 5, 10, 11.

If the scale again does not tip to any side, then the defective ball is number 12 and we can check whether it is heavier or lighter with our last measurement.

If the scale tips to the left side, then either the defective ball is number 9 and is heavier, or it is number 10/11 and is lighter. We measure up balls 10 and 11 against each other and if one of them is lighter than the other, then it is the defective one. If they have the same weight, then ball 9 is the defective one.

If the scale tips to the right side, the procedure is similar.
CASE 2. Let the scale tip to the left side during the first measurement. This means that either one of the balls 1, 2, 3, 4 is defective and it is heavier, or one of the balls 5, 6, 7, 8 is defective and it is lighter. Clearly, balls 9, 10, 11, 12 are all genuine. Next we place balls 1, 2, 5, 6 on one side and balls 3, 7, 9, 10 on the other side.

If the scale tips to the left, then either one of the balls 1, 2 is defective and it is heavier, or ball 8 is defective and lighter. We just measure up balls 1 and 2 against each other and find out which among the three is the defective one.

If the scale tips to the right, the procedure is similar.

If the scale does not tip to any side, then either the defective ball is 4 and it is heavier, or the defective ball is 8 and it is lighter. We just measure up balls 1 and 4 against each other and easily find the defective ball.

The first letter is P if the shapes are circles and B if the shapes are squares.
The second letter is A if the shapes are green and E if the shapes are red.
The third letter is N if the number of shapes is 2 and R if the number of shapes is 3.
The fourth letter is T if the shapes are arranged vertically and S if the shapes are arranged horizontally.
Therefore, the last word should be BENT.

The solutions are shown below.
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The solution is shown below.

1. The stone is sent down in a basket.
2. The son goes down, the stone goes up.
3. The daughter goes down, the son goes up.
4. The stone is thrown back down.
5. The queen goes down, the daughter and the stone go up.
6. The stone goes down.
7. The son goes down, the stone goes up.
8. The daughter goes down, the son goes up.
9. The stone is thrown back down.
10. The son goes down, the stone goes up.

Label the gods with numbers – 1, 2, and 3.
First, ask god 1 “If I ask you whether god 2 is random, would you say ‘Da’?”. If he responds “Da”, then god 3 is not the god of uncertainty. If he responds “Ne”, then god 2 is not the god of uncertainty. In both cases we will be able to find a god which is not the god of uncertainty, let without of generality that is god 3.
Next, ask god 3 “If I ask you whether you are the God of Lie, would you say ‘Da’?”. If he says “Da”, then he is the God of Truth. If he says “No”, then he is the God of Lie.
Finally, ask god 3 whether god 1 is the God of Uncertainty and conclude the identities of all gods.

The solutions are shown below.

Call the 3 liter jug “small” the 5 liter jug “large”.
Fill the large jug with water.
Pour 3 liters of water from the large jug into the small jug.
Empty the small jug with water.
Pour the remaining 2 liters of water from the large jug into the small jug.
Fill the large jug with water.
Pour water from the large jug into the small jug, until the small jug is full.
Now there are exactly 4 liters of water in the large jug.

The answer is Friday.