4a) BCA=90° (angle in semicircle), BCO=90-48=42° (complementary angle).
OA=OB=OC (radii) forming isosceles triangles BOC and COA. ABC=OBC=BCO=42°.
4b) 50m is a common base for two triangles with elevation angle 31°, so distance from the smaller building to the top of the taller is given by the tangent: 50tan31. Distance from the smaller building to the bottom of the taller is 50tan25 (also the height of the smaller building). Total height of the taller building is the sum: 50(tan31+tan25)=53.36m approx.
5a) 10 balls, 2 of which are green. Probability that first ball drawn is green=2/10=1/5. Probability that second ball drawn is green=1/9 (because only one green ball remains and there are 9 balls left). Combined probability is (1/5)(1/9)=1/45.
5b) Since 62 is the average, the total of 8 tests=8×62=496. If one test score is 80, the remaining 7 test scores add up to 496-80=416. If one test score had been 64, the total for all 8 tests would have been 416+64=480 instead of 496. If the new total is 480 for 8 tests, the average m=480/8=60.
I’ve answered all 4 questions so that you can compare your answers with mine.