1 hour and 30 minutes. The wall clock will chime once at 12:15, 12:30, 12:45, 1:00, 1:15, 1:30, 1:45. If you hear 7 consecutive times just 1 chime, then the time is 1:45. If you hear less than that, you will easily find what the time is once again.

It will still take four months (not two).

Turn both hourglasses upside down. When the time of the 7-minute hourglass runs out, turn it upside down. When the time of the 11-minute hourglass runs out, turn the 7-minute hourglass upside down again. Finally, when the time of the 7-minute hourglass runs out, exactly 15 minutes will have been passed.

Since every minute the bacteria double their quantity, the container was half-full at 12:59 PM.

Solve the problem backward. Let the pirates be called A, B, C, D, E, where A is older than B, B is older than C, C is older than D and D is older than E.
If there are only two pirates left – D and E, then the D will keep all the treasure for himself.
If there are three pirates left – C, D, and E, C can propose to give just 1 coin to E and keep the rest for himself. Pirate E will agree because otherwise, he will get nothing.
If there are four pirates left – B, C, D, and E, then B can propose to give just 1 coin to D and keep the rest for himself. Pirate D will agree because otherwise, he will get nothing.
Now if there are five pirates – A, B, C, D and E, A should give coins to at least two other pirates, because otherwise at least three of them will vote negative. Clearly, B will always vote negative, unless he gets offered 100 coins and D will also vote negative, unless he gets 2 coins or more. Pirate A can offer to give one 1 coin to C, 1 coin to E and keep the rest for himself and this is the only optimal proposal – 98:0:1:0:1.

There is just one brother and therefore there are 7 children in the family.

Four – the only number which has the same number of letters as the number itself.

Label the missionaries M1, M2, M3 and the cannibals C1, C2, C3. Then:
1. M1 and C1 cross the river, M1 comes back.
2. C2 and C3 cross the river, C2 comes back.
3. M1 and M2 cross the river, M1 and C1 come back.
4. M1 and M3 cross the river, C3 comes back.
5. C1 and C2 cross the river, C1 comes back.
6. C1 and C3 cross the river.
Now, everyone is on the other side.

Let’s call the people A, B, C, D, E, F. Person A either knows at least 3 among B, C, D, E, F, or does not know at least 3 among B, C, D, E, F.
Assume the first possibility – A knows B, C, D. If B and C know each other, C and D know each other, or B and D know each other, then we find a group of 3 people who know each other. Otherwise, B, C, and D form a group in which no-one knows the others.
If A doesn’t know at least 3 among B, C, D, E, F, the arguments are the same.