Hello, IHATECHONGAS!

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 . . .