What you 'know' 'by quantum mechanics' is not true. Only if you measure an observable which has |0> and |1> as eigenvectors (and not,e.g. |0>+|1> and |0>-|1>) it is true that the states resulting from a measurement are |0> or |1>. Also RANDOMLY suggests not the right thing: the probability for finding |0> may be much larger (or much smaller) than the probability of finding |1>. It depends on the values of a and b in your notation. The art of forcing a state to evolve into a specified one is often termed 'preparation of a state'. Fortunately there are 'selfpreparing states': the ground states for which waiting is often a working mode of preparation.

Just measure the eignevalue associated with the eigenstates |0> and |1>, then if the result is |1>, take another qubit and do it again. Afterward, the state is exactly |0>. The preparation is not conceptually different from the measure.

If you allow an ancillary system, then yes, you can do this. Any measurement (in fact any valid quantum evolution) can be simulated by a unitary evolution of the principal system and an environment, followed by measuring the environment. In your case, you prepare another qubit (call it environment) in state |0>, then evolve the system+environment through a CNOT_{system, environment} followed by a CNOT_{environment, system}, then measure the environment. If you don't have the second CNOT, then you are just measuring the system qubit. The second CNOT ensures that, given result |1>, you are flipping the system qubit back to state |0>.

If you just have access to the system qubit and nothing else, there is no way of "making" the qubit collapse into the desired eigenstate (if this would be true, then quantum mechanics will be no different from classical mechanics).