Is it correct to say that SOME iodine undergoes sublimation at 1 ATM
To answer part 1 of your question:  

It is correct to call the change form solid to gas below the equilibrium vapor pressure sublimation. And there is no separate term for vaporization of a solid above the equilibrium vapor pressure; it is still sublimation.  Similarly, a boiling liquid is still perfectly correctly said to be evaporating.  

To answer part 2 of your question:  

At temperatures below the melting point of iodine, it will sublimate until the equilibrium vapor pressure is reached without going through a liquid phase in that process.  

Once the iodine is heated to it's melting point, it will continue the same process of vaporizing to it's equilibrium vapor pressure. Because it is now a liquid, we just call that evaporation rather than sublimation.  

If the iodine is further heated to it's boiling point at 1 atmosphere, then it will continue to rapidly evaporate as it boils away.  

Regarding the thermodynamic relationships between these processes as mentioned in one of your comments (which I think you should just edit into the question), what you have said is correct. This is a path-independent thermodynamic process, so with respect to the &Delta;H of the phase-change processes, it doesn't matter how you get from the solid phase to the vapor phase. &Delta;H for the direct solid-vapor transition (sublimation) is equal to the &Delta;H of melting plus the &Delta;H of evaporation.  

I hope I have addressed all your points, please ask for clarification in the comments if I've left anything unclear.