Jinkster,
Your first question is addressed in different ways by different schools.
The OATS method is to note that Cdi is proportional to Cl squared.
If speed stays constant then any % increase in weight will cause that% squared increase in induced drag. Fuel flow in a jet is proportional to thrust so the fuel flow will change in proportion to the change in drag.
So in this question we can use (105000/95000)squared = 1.2216. If we assume that all of the drag is increased by this amount then the new fuel flow = 3100 x 1.2216 = 3786.98 (option a).
But all of the above ignores the fact that total drag also includes profile drag, which does not vary with wieght.
The question states that the aircraft are in a holding pattern so we may assume that they are at something close to Vmd (max endurance speed). At this speed induced drag is half of the total drag. So we can say that half of the drag increases by a factor of 1.2216, while the other half stays unchanged.
In effect this will increase total drag by 0.2216/2 = 0.1108. So the total drag (and fuel flow) at the higher weight will be 1.1108 x that at the lower weight. so at 105000 Kg the aircarft will burn 3443 Kg/hr, which is closest to option b.
The alternative (simpler) solution is to consider the total drag and to note that for any small % change in weight at Vmd, there will be approximately that same % change in total darg and jet fuel flow. This gives the simple equation:
FF1/W1 = FF2/W2 for conditions 1 and 2
Rearranging this gives FF2 = FF1 (W2/W1)
Inserting the given numbers gives FF2 = 3100 (105000/95000) = 3426 (option b)
I have discussed this type of question with the examiners who have advised me to "consider all of the drag". So I believe that option b is what they are looking for.
For your second question you need to look at the method and equations on page 12of the CAP 698. I cannot get any of the options exactly, but option b is the closest