Apologies for the delay in further response. I suppose since you are getting for free what I usually charge £25 an hour for I don't really need to apologise
The final one is
not a convergence question, although you are right, the answer is 8.66 degrees, closer to 9. One report of the actual exam question gave it at 64N rather than 60N, giving pretty much 9 degrees as the answer.
Now for technique. The rhumb-line track from (1) to (2) is 090, as is the rhumb-line track (2) to (3). They are all on the same parallel, so the rhumb line between them must be that parallel, which runs East-West.
Draw a diagram with the three meridians (30W, 20W and 10W) and the 60N parallel. The diagram should look basically like a Lambert's chart graticule - meridians converging to the north, parallel a
distinct curve concave to the north. Label WPTs (1), (2) and (3). These are the intersections of the meridians and the parallel.
Join (1) to (2) with a straight line, extending the line beyond (2) but remaining straight (should be south of the parallel beyond (2)). That represents the great-circle track, ans the original question mentions FMS or INS, either of which will direct you on a great circle. Draw a second straight line from (2) to (3), again representing a great circle.
It now becomes obvious that at (2) the aircraft must turn left to fly towards (3). By how much? To turn left onto 090 it would have to turn left by the
conversion angle (CA) between (1) and (2), as the current course is the great circle and 090 the rhumb line - CA is th difference between the two.
CA = 1/2 x ch.long. x sin mean lat = 1/2 x 10 x sin 60 = 4.33
To turn from 090 to the new track is again CA, this time between (2) and (3), as 090 is the rhumb line and the new track the great circle. since the ch.long is the same as is the latitude, this is again 4.33 degrees. So the total left turn is
4.33 + 4.33 = 8.66 degrees
This is far easier with a diagram so if you want a better explanation send me a PM with your email address and I will draw and email one.