Best way to solve these is to put everything into the same context.
So instead of saying the helicopter is flying at 2,500' AGL, say it is flying at 2,850' AMSL. (2,500' + 350').
QFE is 995 hPa, i.e. this is the pressure at the airfield elevation, in other words the pressure at 350' AMSL.
The other aircraft is at FL40, i.e. it is flying 4,000' above the 1013 isobar.
So, helicopter is flying at 2,850' AMSL. Other aircraft flying at 4,000' above 1013 isobar.
Next work out the pressure at sea level. Difference is 350'. Based on 27' per hPa, this would put the pressure at sea level 13 hPa higher, which is 995+13 = 1008 hPa.
We now know the helicopter is flying 2,850' above the 1008 isobar, and the other aircraft at 4,000' above the 1013 isobar. Since 1013 is a higher pressure, it will occur at a lower altitude than the 1008 isobar, this will REDUCE the distance between the two aircraft.
Now it's simply a case of saying how high are each of them above any singular pressure level. Might as well use 1013. There's a 5 hPa difference, which is 135'. Visualising this, it would mean the helicopter is 135' higher than the quoted altitude based on the 1008 pressure level, because the 1013 level occurs below.
So another way of saying the helicopter is 2,850' above the 1008 isobar is that it is 2,985' above the 1013 isobar.
Knowing the other aircraft to be 4,000' above that isobar, subtract one from the other and you get your answer.
1015'.
Hope this helps.
