1. Angle RPB is 72 degrees
2. Arc POQ is 220 degrees.
3. Angle ABC is 135.5 degrees.
Problem 1: Since AR = AB, triangle ARB is isosceles. Also, arc AB is 130 degrees, so angle ARB is 130/2 = 65 degrees, which means angle RAB is also 65 degrees. Then arc RB is 130 degrees. Since arc AQ is 28 degrees, angle RPB is (130 + 28)/2 = 79 degrees.
Problem 2: Since arc POQ is 80 degrees, angle POQ is 160 degrees. Then major arc PQ is 360 - 160 = 200 degrees, so arc POQ on the smaller circle is (360 - 200) + 80 = 240 degrees.
Problem 3: Angle ABX = Angle AOX = 35 degrees, so Angle ABC = 180 - 35 = 145 degrees.
What have you actually tried?
And why is this so "urgent"? Are you just trying to get us to do your homework for you?
First one
Connect RB
Then.....since AR = AB..... then in triangle ARB, angle ARB = angle ABR
And if arc AB = 130 then angle ARB = 130/2 = 65° = angle ABR
So angle RAB = angle PAB = 180 -130 = 50°
And since arc AQ = 28, then angle QBA = angle PBA = 14°
So by the exterior angle theorem angle RPB = angle PAB +angle PBA = 50 + 14 = 64°
Second one
Since in the larger circle, minor arc PQ = 80 degrees
Then central angle POQ also = 80°
But if angle POQ = 80°, then, in the smaller circle..... minor arc PQ = 160°
Then arc POQ = 360 -160 = 200°