Can two electromagnetic fields from different types of sources exist in the same space? Can a high intensity EM field from an electromagnet and an EM field from a high voltage power line occupy the same space? What effects would one have on the other? Would a magnetic field from a solid magnet be able to draw the EM field from a high voltage power line?
Related: https://electronics.stackexchange.com/questions/280558/can-high-voltage-power-lines-provide-a-super-highway-for-drones
 A: (strictly speaking about high voltage power lines)
This is similar to Hall effect, only that you have no additional wires to extract Hall current.
Depending on the direction/strength of the B (magnetic) field it will increase the resistivity of the conductor.
But conduction electrons will not be able to exit the conductor because to get free from the conductor you would need to create such "resistivity" and current that the irradiated frequency is greater than ionization energy of the metal. 
$$ V_{H} = B I l, $$
$$ R_{H} = \frac{V_{H}t}{IB} $$
$$ E = k_B T = hf = R_{H} I^2 dt $$
This is possible but the wire will burn (since temperature from $ k_B T$ is higher than melting point of metal - assuming cables are made of $Al$ or $Cu$) before you could reach such  energy level.
High voltage lines have as low current intensity as possible to avoid losses from $ R I^2 $.
So, from a strong perpendicular magnetic field alone you will heat the conductor, resulting in current loses.
You mentioned something about "drones" ? Well I suppose a small magnet such as in the small electric engines from flying drones (or such as Earth's magnetic field) wont be able create any strong effects.
PS: But if you would have a different setup such as here, with the attachment of additional wires to the high voltage cable, in addition to having the strong magnetic field, you could drain some $V_H$.
Note:


*

*Melting Point of $Al$
$E_{Al}  = 1.38064852e^{-23} * (660.3 + 272.15) = 1.2867e^{-20} Jules $

*Ionization of highest electron (lowest energy) in $Al$
$E_{Al} = 1.6e^{-19} * (5.98 eV) = 9.568e^{-19} Jules $

