First of all, the frequencies are not quantized, as far as no cavity is involved. Given a certain frequancy, e.g. \omega_0. The smallest amount of energy, or in another words, the energy of the single photon which is in such mode (distinguished by the frequency \omega_0, the unit vector of the wave-vector \hat{\mathbf{k}}, and the polarization \epsilon) is \hbar \omega_0.
As you can see, I have assumed that we are talking about light. For other kinds of energy, It's quite the same.
For the light in a particular mode, if it's energy or power is high, that just means the number of the photon in this mode is very large, that's it! We always talking about the pretty week radiation when we use the quantum theory. so if only one photon in this mode, the energy is \hbar \omega_0. Two photons, 2\hbar \omega_0,..., then we have a series of energy level.
For a "real radiation", like a pulse or something, they are composed by a lot of plane waves (Fourier transformation). Each plane wave can be treat as a mode. So, yes, the frequency is continuous. But for each mode, the energy is quantized. To get the energy of the pulse, you need to integral over the whole modes of the pulse (loosely, over all the frequencies, if the polarization is not considered). As long as the total energy is low, it is still quantized, since adding one photon into a particular mode of the pulse can change the total energy dramatically. On the other hand, if the energy is high. it should be treated as a continuous quantity.
If the spectrum of the pulse is wide, this also means that the pulse is composed by a lot of photons (most of which are in different modes), so generally, the energy, or power of the pulse can not be too low, then it is continuous. Adding one photon into the pulse won't change it too much.