Genome is correct where he has answered
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You will not be asked to calculate the great-circle distance except in cases around the equator or along meridians. The general formula for a great-circle distance is specifically excluded from the syllabus, so if the points are not on the equator and not on the same meridian (or a meridian and its anti-meridian, for which the great-circle route is via the pole) then the question is not about calculating the great-circle distance.
Note you may be asked to measure a distance. Measure on the appendix given, don't try to calculate it as some did.
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There is no such term as 'convergence angle', this is a source of much confusion. You must distinguish between 'convergence' and 'conversion angle'. Convergence is the angle between two meridians, giving the change of bearing along a great-circle arc (for Earth convergence) or along a straight line on a chart (chart convergence). Conversion angle (C.A.) is the approximate angle between the shorter great-circle arc and the shortest rhumb line between two points.
This question requires you to recognise that flying a rhumb line between these three points the bearing would be a constant 090 degrees. Draw a diagram it will show that the great-circle track (1) to (2) measured at (2) is less than 090 degrees by the C.A. between (1) and (2), about 4.5 degrees.
The track from (2) to (3) measured at (2) is less than 090 degrees by 4.5 degrees. Therefore at (2) the aircraft track changes from 090 + 4.5 to 090 - 4.5, a left turn of 9 degrees.
This is hard to see without the diagram, if you're having problems getting that right then send me a PM with an email address and I'll send you a diagram.
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20 kts
A couple of ways to do this one. The simplest way (but it only works for a given point of slowing, not for a given new airspeed where you have to work out the point at which to slow down) is to calculate the time taken, then add the delay required and recalculate the speed from the distance and the new time. At 265 kts, 230 kts G/S this is about 38 minutes. To delay by 5 minutes then the plan is to take 43 minutes. This requires a groundspeed of 210 kts, which will require a drop in speed of 20 knots.
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Solar system
66°, of those choices. Actually it's somewhat higher, as the centre of the sun reaches the celestial horizon every day up to about 66°30' (maximum declination of the sun) but for sunrise only the top need be seen above the visible horizon (lower than the celestial due to atmospheric refraction), so that applies at a higher latitude.
Hope this helps
Send Clowns
Gen Nav Instructor
Bournemouth Commercial Flight Training