In case a), you are correct.The slip is zero so no current is induced in the rotor. The stator current is the magnetizing current. The power drawn from the source will be the stator iron losses. This condition is almost the same as no-load operation, but driving the rotor to exactly synchronous speed will mean that the the mechanical losses, friction and aerodynamic drag (windage) will be supplied by the external drive.
Case b) is called plug braking, the motor brakes the load with both the braking energy and high rotor slip losses dissipated in the rotor. The stator current will be somewhat higher than the motor's locked rotor current.
Re case b) effect on motor:
The image below shows torque power and current vs. slip for a particular induction motor. At slip = 2, the motor's braking torque is about 66% of rated torque and the braking power (P = -5.3kW) is about 66% of rated motor power. The copper losses in the rotor are about 10.8 kW, making the total power dissipated in the rotor 16.1 kW, about 65 times the power dissipated during normal full load operation. The copper losses in the stator at s=2 are about 22 kW, about 44 times the power dissipated during normal full load operation. When the motor is started at full voltage, the power dissipated in the rotor and stator is initially (S=1, locked rotor condition) about 39 times tor normal power dissipation in each. Either locked-rotor or S=2 operation will cause the motor to fail due to internal heating rather quickly. For S=2, the heating is more severe, particularly in the rotor.
Image and motor data from Fitzgerald, Kingsley, Umans Electric Machinery 4th ed.