Solution:

2nd

When the cylinder is spun for the first and last time, the following six outcomes are possible:

First:
First chamber is fired first. The first player loses.

Second:
Second chamber is fired first. The first player loses.

Third:
Third chamber is fired first. The first player loses.

Fourth:
Fourth chamber is fired first. The second player loses. (First shot, player 1, chamber 4 empty. Second shot player 2, chamber 5, empty. Third shot player 1, chamber 6 empty. Fourth shot player 2, chamber 1 not empty.)

Fifth:
Fifth chamber is fired first. The first player loses. (First shot, player 1, chamber 5 empty. Second shot player 2, chamber 6, empty. Third shot player 1, chamber 1 not empty.)

Sixth:
Sixth chamber is fired first. The second player loses. (First shot, player 1, chamber 6 empty. Second shot, player 2, chamber 1, not empty)

Thus, if you analyze, you will find that the second player has a 4/6 chance of winning. Therefore if you want to win the game, going second gives you better shot at it.

Solution:

16 (5 + 1 + 1 + 4 + 5)

black coloured squares : 5
red coloured square : 1
blue coloured square : 1
small green coloured squares in middle : 4
large green coloured squares : 5

Solution:

Like there is no tomorrow.

Solution:

The time at first observation:
1:12

The time at second observation:
3:24

Thus, two hours and twelve minutes have passed between the two observations.

Solution:

18

Solution:

end up behind bars

Check that the word "END" is spelled in an upward direction behind three occurrences of the word "BAR".

Solution:

Cousins

Explanation:
Parent of Beta is Alpha.
Parent of Zeta is Epsilon.
Parent of Delta is Gamma.

Ergo, Epsilon is Alpha's brother and Gamma is Alpha's sister.

Therefore, Zeta, Beta and Delta are cousins.

Solution:

Mixing business with pleasure.(PbLuEsAiSnUeRsEs)

Solution:

She has misplaced 119 pens in total.

Solution:

To make things easier, let us just denote each box with the help of R number and C number i.e. the row number and the column number.

The rule that has been used here is:
R4C1 = (R1C1 * R3C1) - R2C1 = (5 * 2) - 8 = 2
R4C2 = (R1C2 * R3C2) - R2C2 = (7 * 4) - 10 = 18
R4C3 = (R1C3 * R3C3) - R2C3 = (6 * 5) - 12 = 18

Using the same rule for the last box:
R4C4 = (R1C4 * R3C4) - R2C4 = (4 * 9) - 17 = 19

Thus 19 is the missing number.