How does the centripetal force vary with the radius of the circular path? Consider (a) constant frequency a?

Fc = m*Ac = m*V²/R = m*R*w².....Both your questions can be answered with these 2 eqs.

No.  There is no displacement in the radial direction, so no work can be done.

Constant frequency is F = W/2pi where W is a constant angular speed.  And if you don't like angular speed V = WR is a constant tangential speed (linear) when the radius is fixed.

We find centrifugal force C = mV^2/R = mW^2 R = m (2piF)^2 R = P centripetal force when R is fixed.

So when R is fixed and F is increased, centripetal force increases.  EX: double F and P quadruples.

When both can vary, we have dP = m 4pi^2 [ 2F R dF + F^2 dR].  As you can see, we can't generalize here as dP = 0 can be achieved when  [ 2F R dF + F^2 dR] = 0 so that dF/F = - dR/R * 1/2 and dF/F + 1/2 * dR/R = 0  In other words, the only way we can specify how the centripetal force is altered when both are free to vary is to put in concrete values for F and R.  

EX:  A dF/F = 10% increase in frequency would be offset by a 20% decrease in radius; so dP = 0 = 10% - 1/2 *20% would result and there would be no change in the centripetal force despite changing both the frequency and the radius.