how would inductor react if I inject current impulse into parallel LC tank?
Injecting a current impulse onto a L-C parallel tank is like hitting a bell with a hammer.  If nothing was going on before (voltages and currents were 0), then this will start the tank ringing.

The inductor can't change it's current instantly, but the capacitor can (I'm assuming we're talking theoretical ideal components here).  The impulse will have no immediate affect on the inductor's state.

A current impulse on a capacitor causes a step change in voltage.  Depending on whether this step adds or subtracts from the existing voltage, it either adds or subtracts energy from the system.

For example, if you happen to hit the system with +1 V when the capacitor has -1 V on it and the current is 0, then you've just removed all the energy from the system and now everything will just sit there at 0.  On the other hand, if the voltage was +1 V and the current 0, you've now doubled the voltage and quadrupled the energy of the system.  The amplitudes of the voltage and current sines will be twice what they were before.