For #1, extend QRS across the circle to meet the other side at T and drop a perpendicular from the centre O onto this line meeting it at N.
TQ.RQ = PQ^2, so (TR + 3).3 = 36, so TR = 9.
NS = 9/2 - 3 = 3/2.
From the triangle ONS, use Pythagoras to calculate ON^2, and then from the triangle ONT use Pythagoras again to calculate OT, the radius of the circle.
There's a trig method for #2.
Join M to B and A.
Since angle C is a right angle, BA, (length 10) will be a diameter of the circle and in which case angle BMA is also a right angle.
Angles ABM and BAM are both equal to 45 deg, (angles on the same arc), so BM = AM.
Now use Pythagoras in the triangle ABM, BM^2 + AM^2 = 2BM^2 = 10^2 = 100, BM = sqrt(50) = 5sqrt(2).
Call angle CBA theta, then using the sine rule in triangle CBM,
5sqrt(2)/ sin(45) = CM/ sin(theta + 45),
so CM = 5sqrt(2)(sin(theta).cos(45) + cos(theta).sin(45))/sin(45).
Simplify that to get 7sqrt(2), (same as geno).