Do photons with a frequency of less than 1 Hz exist? A photon with a frequency of less than 1 Hz would have an energy below
$$
E = h\nu < 6.626×10^{−34} \;\rm J 
$$
which would be less than the value of Planck's constant. Do photons with such a low energy exist and how could they be detected? Or does Planck's constant give a limit on the amount of energy that is necessary to create a single photon? 
 A: The Planck constant is $h=6.626\times 10^{-34}\;\rm J\,s$ (joule seconds), you cannot compare it to an energy which is measured in Joules - this is the flaw in your argument. To answer the question: Such low energy photons can exist in principle, however the question is how to actually generate them. I'd propose to take a possibly low energy photon and redshift it (check Doppler effect). It will, however, be in very red radio range and therefore hard to detect.
A: The frequency $\nu$ is in seconds$^{-1}$, which is purely human-based unit having a relation to rotation of the Earth. Thus no reason why 1 Hz was a limit. Planck $h$ value is also not massless unit and it's value has relation to SI system.
Existence: while I don't see a principal reason for non-existence of such a photon, neither I see a physical process, that would generate such a radiation. You would need some anthena of length $\sim 10 ^8$ meters and some process that would correlate a charge across such a distance.
A: You don't need a long antenna to radiate at 1 Hz.  You need a long antenna to radiate efficiently at 1 Hz.  The efficiency is proportional to the cube of the length of the antenna in wavelengths (look up {electrically} 'small antennas').  A big 1 Hz current in a short wire will radiate very little power, but 1 photon a second would be 6.6e-34 Watts, so the numbers may be in favour of radiation.  1 Hz ==> 3e8 m wavelength, so 1 m long wire antenna may have efficiency of order 3e-26, which looks like lots of photons per watt into the antenna (most of the watt goes into resistive, dielectric and magnetic losses in the matching circuit or the generator).
A: The shortest answer for why such photons "exist" is that whether a given photon qualifies depends on your rest frame. Take your favourite high-frequency photon in the universe, say of frequency $n$ Hertz in your rest frame. With a Lorentz boost $-\beta$, I multiply this by $\gamma (1-\beta)=\cosh\phi-\sinh\phi=e^(-\phi)$ with $\phi$ the rapidity $\phi=\mathrm{artanh}\beta$. Setting $\phi>\ln n$ (equivalently, $\beta>\frac{1-n^{-2}}{1+n^{-2}}$) does the job.
A: Yes. Essentially any frequency > 0 is theoretically possible. You may have confused this with the concept that it’s not possible for an electromagnetic wave with a given frequency f to have an energy less than E = h.f without eliminating the entire wave.
