@Hyeon_Woo_Kim glad my answer helped!
To your second question, this is again related to the normalization of compositions, which affects how you balance the equation you wrote. The equation you describe for E_above_hull is actually not the energy above hull (which would be zero) but similar to the equation for the reaction enthalpy, \Delta H_{\textrm{rxn}} for the reaction \ce{Rb2O + 2 InO3 -> Rb2In4O7}. The reaction enthalpy for this balanced reaction should be -1.286 eV or -0.0989 eV/atom, which is actually the number in the green box on the image in your first post.
The number you calculated (-0.188), should match this per-atom number if you repeat the calculation after ensuring that it is stoichiometrically balanced. This could be done using the reaction above, or by multiplying the term in parentheses (E_{\ce{Rb2O}}*0.33 + E_{\ce{In2O3}}*0.667) by a factor of 3. Once you calculate the energy from this reaction, you should divide by the total number of atoms being reacted (13 atoms) to get the -0.0989 eV/atom number.
The -0.429 eV/atom number that is in the compound phase diagram you plotted is somewhat enigmatic, because it is the result of setting the normalization to False. I think this means that the energy is calculated from a non-balanced reaction, so perhaps it should not be used…