You leave with 2 cakes. Every time you cross a bridge, you give one of them to a troll, and then get it back. Eventually, you will arrive at your grandma with exactly 2 cakes.

He needs only 21 wrappers. He uses them to get 7 chocolates. After he eats them, he is left with 7 wrappers, which uses to get 2 more chocolates. After he eats these, he is left with 3 wrappers, which are enough to get one more chocolate and make them 10 in total.

If all four octopuses lied, then their total number of legs must be 28, which is impossible. Therefore exactly one of them said the truth, and the other three had 7 legs each. Since the truthful octopus must have 6 or 8 legs, and 21 + 8 = 29, we see that it has exactly 6 legs, and therefore it is the second one.

First, we note that if we have 1 gold ball only, then we need:
1 measurement in a group of 2 balls
2 measurements in a group of 4 balls
3 measurements in a group of 8 balls
Start by measuring 1, 2, 3, 4, 5.
If there are gold balls in the group, then measure 6, 7, 8, 9, 10, 11.
If there are gold balls in the group, then measure 5, 6, 7.
If there are no gold balls among them, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 8, 9, 10, 11, so we can find the gold balls with the remaining 2 measurements.
If there are gold balls in 5, 6, 7, then measure 5, 8, 9. If there are gold balls there, then 5 must be gold, and we can find the other gold ball among 6, 7, 8, 9, 10, 11 with the remaining 3 measurements. If there is no gold ball among 5, 8, 9, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 6, 7, so again we can find them with only 3 measurements.
If there are no gold balls in the group, then measure 5, 12, 13.
If there are no gold balls among them, then measure 14, 15. If none of them is gold, then measure individually 1, 2, and 3 to find which are the 2 gold balls among 1, 2, 3, 4. Otherwise, there is a gold ball among 1, 2, 3, 4, and among 14, 15, and we can find them with the remaining 3 measurements.
If there are gold balls among 5, 12, 13, then measure 5, 14, 15. If none of them is gold, then there is a gold ball among 1, 2, 3, 4, and a gold ball among 12, 13, so we can find them with 3 measurements. Otherwise, 5 is gold, and again we can find the other gold ball among 1, 2, 3, 4, 12, 13, 14, 15 with 3 measurements.
If there are no gold balls among 1, 2, 3, 4, 5, then we measure 6, 7, 8.
If there are gold balls in the group, then measure 9, 10, 11, 12, 13.
If there are no gold balls among them, we measure individually 6, 7, 8, 14.
If there is a gold ball among 9, 10, 11, 12, 13, then there is another one among 6, 7, 8. We measure 8, 9. If none of them is gold, then we can find the gold among 6, 7, and the gold among 10, 11, 12, 13, with 3 measurements total. If there is a gold ball among 8, 9, then we measure 10, 11, 12, 13. If none of them is gold, then 9 is gold and we find the other gold ball among 6, 7, 8 with 2 more measurements. If there is a gold ball among 10, 11, 12, 13, then we can find it with 2 measurements. The other gold ball must be 8.
If there are no gold balls in the group, then measure 9, 10.
If there are no gold balls among them, then measure individually 11, 12, 13, 14.
If there are gold balls among 9, 10, then measure 11, 12, 13, 14. If there is a gold ball among them, then there is another one among 9, 10, and we can find them both with 3 measurements. Otherwise, we measure 9 and 10 individually.

The numbers are 2, 3, and 5. First, we check that these numbers work.
Indeed, Frank would not be able to figure out whether his number is 2 or 8. Then, Gary would not be able to figure out whether his number is 3 or 7, since with numbers 2, 7, 5, Frank still would not have been able to figure his number out. Finally, Henry can conclude that his number is 5, because if it was 1, then Gary would have been able to conclude that his number is 3, due to Frank’s inability to figure his number out.
Next, we we check that there are no other solutions. We note that if the numbers are 1, 4, 3, or 3, 2, 1, or 4, 1, 3, neither Frank nor Gary would have been able to figure their number out. Therefore, if the numbers were 1, 4, 5, or 3, 2, 5, or 4, 1, 5, Henry would not have been able to figure his number out. Thus, 5 is not the largest number.
Similarly, if the numbers are X, X + 5, X + 10, or X + 5, X, X + 10, once again, neither Frank nor Gary would have been able to figure their number out. Therefore, if the numbers were X, X + 5, 5, or X + 5, X, 5, Henry would not have been able to figure his number out.

A lost it. Since there have been (10 + 15 + 17) / 2 = 21 games played in total, and each player never misses 2 games in a row, the only way for A to play just 10 games is if he plays the 2nd, 4th, 6th, etc. games, and every time loses.

Coming soon.

The total of all numbers on the clock is 1 + 2 + … + 12 = 78. Therefore each piece must contain numbers with total sum 26. The only way for this to happen is if the pieces are broken via 3 parallel lines – {1, 2, 11, 12}, {3, 4, 9, 10}, {5, 6, 7, 8}.

The Rubik’s Chess puzzle cannot be solved. You can see a detailed solution HERE.

They need 17 minutes. Label the friends A (1min), B (2min), C (7min), D (10min). A and B cross the bridge, then A returns back with the flashlight. C and D cross the bridge, then B returns back with the flashlight. Finally, A and B cross the bridge.
In order to see that this is optimal, notice that when D crosses, he needs at least 10 minutes. If C crosses separately, this will make already 17 minutes in total. Therefore C and D must cross together, and A and B must be at that time on the two opposite sides of the bridge. From here it is easy to conclude that the friends indeed need at least 17 minutes.