The solution is shown below.

The person is a baseball player. The men are a catcher and an umpire.

The solution is shown below.

Keep up adding the given numbers and remember only the last two digits of the sum. In the end, if the result is less than 50, subtract it from 50. If the result is larger than 50, subtract it from 150. This method works because the sum of all numbers from 1 to 100 is 5050, so if you know the sum of all the listed numbers, you will know the missing number as well.

The secret of the trick is to memorize the group of cards which are located in even positions and the group of cards which are located in odd positions in the original deck. An easy way for doing this is to split the cards into two groups, such that the cards in the first group are only spades and diamonds, and the cards in the second group are only clubs and hearts.
When the two assistants pick their cards and then return them back into the deck, the order of the cards is reversed. When you split the original deck into two piles (even after cutting it several times), each of the piles will contain a card which should not be there. For example, the group of spades and diamonds will contain one clubs card, and the group of clubs and hearts will contain one diamonds card. These two cards are the ones which were picked by the assistants.

No, the spiders will always be able to contain the beetle within the carpet. We draw two perpendicular lines passing through the beetle which are parallel to the diagonals of the square. The spiders’ strategy is to follow the four points where these lines intersect with the boundary of the square. When a spider’s corresponding intersection point moves to an edge of a different color, the spider waits in the corner. In order to accomplish this strategy, the speeds of the spiders need to be at least √2~1.4 times higher than the speed of the beetle.

The second programmer wrote down “1” in front of the first circle. When the second programmer lifted his hand, he saw the number “10…00”, which is exactly the number of possible head-tail combinations in binary system.

The shortest game takes just 4 moves:
1. Nf3 e5
2. Nxe5 Ne7
3. Nxd7 Nec6
4. Nxb8 Nxb8

Yes, you can, as shown on the image below.

The best move is 1. Rc8+. Then either 1. … Kxb7 2. Rxd8, or 1. … Rxc8 2. Qxa7 Kxa7 3. bxc8 = N+, and Black loses both the Rook and the Queen.