Well, I finally pull it out.
I used Green's functions and it was pretty straightforward,
For a harmonic oscillator, you have to solve:
$(\frac{d^2}{dt^2} + 2b\frac{d}{dt} + \omega^2)G(t-t')= \delta(t-t')$
The solution is for $t>t' $:
$$
G(t-t')= exp(-b(t-t'))\frac{\sin(\omega'(t-t'))}{\omega'}
$$
where $\omega' = \sqrt{\omega^2-b^2}$
The solution is:
$$
y(t)= \int{f(t')exp(-b(t-t'))\frac{\sin(\omega'(t-t'))}{\omega'}dt'}
$$
Using $f(t) = \sum{\delta(t-nT)}$ the integral becomes super easy and you can interchange the sum and the integral since the sum does not depend on t':
Finally:
$$
y(t)= \sum{exp(-b(t-nT))\frac{\sin(\omega'(t-nT))}{\omega'}}
$$
So what we got is as many sine functions as delta diracs the comb has, and vibrating at the natural frequency (just like a guitar) regardless if you are plucking it with a determinated frequency.
The power transfer is maximised at resonance because the driving force and the velocity of the oscillator are in phase.
If you multiply two sinusoidal terms together (the force and the velocity) with a phase difference between them, then the product has its maximum average value when the phase difference is zero and a minimum value when the phase difference is $\pm \pi/2$.
Your steady state solution could be correct, but it is more usual to say that if the driving force is $F_0 \sin \omega t$, then the displacement $x \propto \sin(\omega t + \phi)$, where the phase difference $\phi$ is given by
$$ \phi = \tan^{-1}\left(\dfrac{-\gamma \omega}{\omega_0^{2} - \omega^{2}}\right),$$
and $\gamma$ is the damping coefficient.
You can see that when $\omega = \omega_0$ the phase difference between displacement and force is $-\pi/2$. But if you differentiate the displacement to get the velocity $$v \propto \cos(\omega t + \phi) = \sin(\omega t + \phi +\pi/2)$$
and at resonance the phase difference between velocity and force is zero.
If the power transfer is maximised, then this is also why the amplitude is maximised, since the velocity amplitude also increases with the amplitude of the displacement.
Best Answer
When you have an overdamped oscillator and release the mass from rest with some deflection, it will return to zero without ever crossing (no oscillation). For a lightly damped oscillator, you would see oscillations (zero crossings) whose amplitude becomes smaller with time.
Now if your mass is launched towards zero (the equilibrium) with sufficient speed, it will cross zero (once) before the damping slows it down. The question asks about the maximum speed that just doesn't cause a zero crossing (and which therefore "throws" the mass to the equilibrium position in the shortest time, most likely).
A diagram might help to explain: