Regarding section (c)(i),
I will try to help by offering an analogy; imagine someone standing in the rain with an umbrella, the rain falls straight down and the umbrella is held conventionally with the protective surface horizontal. If you were to ask how much rain is hitting the umbrella it should be easy to see that the area of the umbrella is directly related to the amount of water that will hit. If you double the area of the umbrella you will double the amount of water making contact with it. Also, if we are asking how much rain will hit in a certain amount of time and if we assume the density of raindrops to be constant, then we will need to know the velocity of the rainfall. You might be able to see that the amount of water hitting the umbrella in a certain amount of time is contained in a cylindrical column of water with an end area equal to the umbrella’s area and the height given by the velocity of rainfall times the time we are concerned with.
To relate this to your problem, you are given the area swept out by the blades and the velocity of the wind. The density of air is constant and the problem is concerned with a 1 second time period.
“...the mark-scheme and they said mass = Area * velocity...”
Technically speaking this is incorrect.
The correct statement is mass=density*volume=density*area* length=density*area*velocity* time
The “mark scheme” left out density and time. In this case, since the time is 1 sec, the numerical value of time can be “left out” since it is 1, leaving density*area*velocity.
I hope this makes sense and is helpful.