How to predict SN2 reaction when comparing nucleophilicty and leaving group ability
This is very famous "Finkelstein Reaction" that involved the exchange of one halogen to another through SN2 mechanism. The driving force for the reaction is the solubility of salts (NaI, NaBr, KF) under equilibrium. 
There are different factor that effect the equilibrium position of a reaction, firstly, nucleophilicity of anion, presence of  good leaving group, and stabilization of anion in a given solvent. 

first part of your question, I don't think with the absence of sodium cation the reaction would take place, because the driving force is the precipitation of salts of NaBr and NaCl in acetone.

In the second part of your question, you are right iodide is better leaving group compared to Bromide because iodide is weaker base compared to bromoide. Apparently, iodide can't replace bromine but here the reaction is driven toward products by "mass action" due to the precipitation of the insoluble salt as mentioned above.