Do the compounds NaₓIrO₂ and NaOsO₂ exist? Based on some physics background, I want to know whether have scientists synthesized these two materials: $\ce{Na_{$x$}IrO2}$ and $\ce{NaOsO2}$ in the lab, if so, what properties do they have? 

I will be very appreciative if someone could recommend me some relevant articles. 


  Would you be willing to share your interest in these two materials? For example, is there a specific application in mind? If so, we might be able to suggest alternate materials should these two not exist.


I want to seek a material that has triangular lattice structure, strong spin-orbital coupling (e.g. 5d element) and large (or moderate) coulomb interaction between electrons. For example, $\ce{ABO2}$ type, $\ce{NaOsO2}$ compounds, where $\ce{A}$ and $\ce{B}$ are alkali and transition-metal ions, respectively. So far, I know $\ce{IrTe2}$ may be a one. 
Do you mean that both $\ce{NaxIrO2}$ and $\ce{NaOsO2}$ do not exist?


  Is there any information on iridium and osmium oxides in these low oxidation states?


For example, $\ce{Na2IrO3}$ and $\ce{Li2IrO3}$. But these two belong to the honeycomb lattice structure, and what about the triangular lattice structure?
For $\ce{Na_{x}IrO2}$ the current answer seems to be "no" it has not been made but it is discussed in the literature theoretically.

From Kitaev-Heisenberg models for iridates on the triangular, hyperkagome, kagome, fcc, and pyrochlore lattices:


  However, no triangular lattice iridate with the relevant edge sharing octahedra structure is currently available; a compound of the
  type NaxIrO2 may or may not turn out to be stabilized.


and in Chiral spin liquid and topological superconductivity in the undoped and doped
triangular Kitaev-Heisenberg model: 


  In addition, a class of ABO2 (where A and B are alkali and transition metal ions, respectively) type transition metal compounds and the possible material NaxIrO2 may be potential candidates for realizing the triangular Kitaev-Heisenberg model.


(But that's your paper isn't it? I don't have any information beyond what you already know, so really this is a useless answer.)