Hey, thats a nice clear diagram. You will need some basic circle geometry to solve these questions.
First draw a line from point O to T. Where it intersects the line AB I will call point P.
Now using circle geometry, we know line OP bisects the line AB at 90 degrees (I don't remember the exact wording of the theorem but look it up if you need to).
So you have a right angled triangle OPA with right angle OPA, hypothenuse 5, and height (9/2) = 4.5
Let angle OAB = x. Cosine (x) = (adjacent side)/hypothenuse = 4.5/5
so x = arcsin (4.5/5)
You actually don't need to use trigonometry here. Pythagoras equation will work with a = PA and c = OA.
But if you want to use trig, from the previous working, you have angle OAB = x.
Sine (x) = (opposite side) / hypothenuse. You have hypothenuse = OA = 5, sine (x) you can calculate using the value of x you got from Qa. Now you can solve for opposite side which is perp distance O to AB.
See if you can do this yourself, ask for help if you need it. Note that the tangent line is always perpendicular to the centre of the circle (i.e. right angles)