For the last two, there is a theorem.
From an external point, draw two secants to a circle
. . (or two tangents or a sectant and a tangent).
The angle formed is one-half the difference of the intercepted arcs.
In #2, we have a secant and a tangent forming a 20° angle.
. . The intercepted arcs are 60° and x.
So we have: .½(x - 60) .= .20 . → . x = 100
In #3, we have two tangents forming a 52° angle.
. . The intercepted arcs are x and 360-x.
You finish it . . .