There is quite a lot of explanation that should accompany these answers. I hope you've studied your circle theorems.
(a) because you were given that , which is half of the intercepted arc BC.
(b) because you can find the measure of arc AE by subtracting the other known arcs from 360. Arc AE measures 80 degrees. That makes the inscribed angle = 40.
(c) because the angle formed by the intersection of two chords is one-half the sum of the measures of the two subtended arcs. Arc AE = 80 degrees and arc BC = 100 degrees. Their sum is 180, and half that is 90 degrees.
(d) because the exterior angle formed by two secants is equal to one-half the difference of its intercepted arcs. Thus angle F is 1/2(150 - 80) or 35 degrees.
(e) since it is in the right triangle FAD and is complementary to angle F.
Your second question requires even more explanation. I suggest you post it separately and tell us how far along you have gotten. Be specific about what you don't understand.
Study your circle theorems (all about central angles, inscribed angles, tangents, secants, chords, arcs, etc.).
I have a problem with only two of your answers.
(c) because angle CPT is supplementary to angle CPB. Now we know that in triangle CBP, angle ACB is 80 degrees. We found that using the proportions. Angle CBP is 30 degrees because angle CBA was found to be 60 degrees using the proportions, and according to the given, it is bisected by BT. So, in triangle CBP, we have angles of 30 and 80 degrees. That leaves angle CPB to be 70 degrees. Angle CPT is supplementary to angle CPB because they make up a linear pair.
(e) because it is one-half the difference of the intercepted arc measues on the circle. This angle intercepts arc AX (not labeled in diagram) and arc AB. Arc AX measues 60 degrees because the intercepted angle ABP = 30 degrees. Arc AB measures 160 degrees since its inscribed angle ACB measures 80 degees. So, angle T = 1/2(160 - 60) = 50 degrees.