Solution:

In the first wave, let us make an assumption that there is only 1 red hat and other 8 hats are blue. The student who can see all the other 8 blue hats on others' head will certainly be clear that there are 8 blue hats and 1 red hat since the professor told them that there is at least 1 red hat among the nine. But no one was able to answer after the first half an hour. Thus this can't be possible and our assumption is wrong.

Now let us assume that there are 2 red hats and 7 blue hats. The student who is having a red hat on his head can definitely see 7 blue hats and one red hat. He can deduce that he is wearing a red hat. But nobody was able to answer in the second interval as well. Thus our assumption is again wrong.

In the last wave of final five minutes, let us assume that there are 3 hats and 6 blue hats. The student with a red hat can see 2 other red hats and 6 blue hats and he may deduce that he is wearing a red hat. In this interval, everyone was able to answer which states that our assumption is correct.

Thus there are 3 red hats and 6 blue hats.

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

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:

The Lady sitting in the chair has the detector attached to the back of her blouse.

Solution:

Since the knife was in her hand which implies that she was planning to kill her husband "Inspector Montalbano".

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.

Solution:

8 liters jar: 8; 5 liters jar: 0; 3 liters jar: 0
Fill 5 liters jar entirely.
8 liters jar: 3; 5 liters jar: 5; 3 liters jar: 0
Fill 3 liters jar with 5 liters jar.
8 liters jar: 3; 5 liters jar: 2; 3 liters jar: 3
Pour entirely from 3 liters jar to 8 liters jar.
8 liters jar: 6; 5 liters jar: 2; 3 liters jar: 0
Pour entirely from 5 liters jar to 3 liters jar.
8 liters jar: 6; 5 liters jar: 0; 3 liters jar: 2
Fill 5 liters jar with water from 8 liters jar.
8 liters jar: 1; 5 liters jar: 5; 3 liters jar: 2
Fill the 3 liters jar with water from the 5 liters jar.
8 liters jar: 1; 5 liters jar: 4; 3 liters jar: 3
Empty 3 liters jar in 8 liters jar.
8 liters jar: 4; 5 liters jar: 4; 3 liters jar: 0

Now you have 4 liters of water in 8 liters jar as well as 5 liters jar.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

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:

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:

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.