(a) Apply Lenz's law, induced current will flow to oppose the magnet. So South-pole is developed at end $\mathrm{q}$ and current flows in qrpq direction.
(b) Same as above, South-pole developed at $\mathrm{q}$ and current is in prqp direction. For coil 2, North- is going away so attractive i.e., South pole is developed at end $x$ and direction is yzxy.
(c) As key is closed, magnetic flux rises in the first coil, in other coil, induced current would be such as to oppose this increasing magnetic flux. This happenes when magnetic field produced in this coil is from right to left hence current is in yzx direction.
(d) As rheostat is changed to decrease resistance, current will increase and magnetic flux linked with it will rise. Applying Lenz's law, the induced current would oppose this increase, hence direction of magnetic field should be right to left and current is in direction $z y x$.
(e) When current is flowing before release of tapping key magnetic flux is from left to right (right end is $\mathrm{N}$ pole). As key is released, current decreases magnetic flux decreases, hence induced flux would oppose this decrease and favour an increase in flux. Hence direction of induced current in adjoining coil is from left to right in direction xry.
(f) As current decreases in straight conductor, magnetic field in the plane of the coil remains to be in the same plane. The circular coil does not see the change in magnetic field (same plane) hence no current (or flux) is induced.