1. The key here is recognizing that you can burn a fuse at both ends and also burn two fuses at the same time. Burning a fuse at both ends means that it'll burn twice as fast. That means instead of burning one hour, by burning on both ends, it'll burn only half an hour.
You can solve this problem by burning one fuse on one end, and another on both ends. When the one burning on both ends finishes, then you know that 30 minutes have passed. At this time, you know that the fuse only burning on one end will burn for exactly 30 more minutes. Light the other end and voila, it'll finish burning in 15 minutes instead of 30. That means you get exactly three-quarters of an hour when the second fuse finish burning.
2. This is just algebra. The variable you're trying to solve is the number of years Diophantus lived. Let's call this X. The other variable is the number of years his son lived. Let's call this Y.
You know that Y = 1/2 X.
Then you can figure out the length of time the son lived that overlapped with X.
You know the son is born 1/6X + 1/12X + 1/7X + 5 (which is 11/28X + 5)
You know the son died 1/2X later.
When the son is born, Diophantus has (17/28X - 5) years remaining in his life.
You also know that the remainder of his life is really just his son's life plus 4 years.
So you have two expressions basically describing the remainder of Diophantus's life. They are equal because they describe the same thing. Therefore:
17/28X - 5 = Y + 4
But you know that Y = 1/2X, so now you have two equations and two variables and you can solve for X. So 51/84X = 42/84X + 9
X would be 84.