The isomorphism (1) comes from the fact that the systems $((I+J)^n)_n$ and $(I^n+J^n)_n$ define the same topology on $A$: $ I^n+J^n \subseteq (I+J)^n, \quad (I+J)^{2n} \subseteq I^n+J^n.$
Isomorphism (2) is more complicated (at least I don't have a simple proof). Fix an $m$. For any $n\ge m$, there is a canonical exact sequence $ 0\to J^m/(J^m\cap (I^n+J^n)) \to A/(I^n+J^n)\to A/(I^n+J^m)\to 0.$ Passing to the limit, we get a canonical exact sequence $ 0 \to \varprojlim_n J^m/(J^m\cap (I^n+J^n)) \to \hat{A}:=\varprojlim_n A/(I^n+J^n)\to A_m:=\varprojlim_n A/(I^n+J^m)\to 0.$ The exactness on the right comes from the surjectivity of $J^m/(J^m\cap (I^{n+1}+J^{n+1}))\to J^m/(J^m\cap (I^n+J^n)).$ As above, the $J^m\cap (I^n+J^n)$ define the same topology as the $J^m\cap (I+J)^n$. By Artin-Rees lemma, the latter define the same topology as the $J^m(I+J)^n$. So the lefthand side term in the above exact sequence is $(J^m)^{\hat{}}=J^m \hat{A}$. Therefore $A_m\simeq \hat{A}/J^m \hat{A}$. Passing to the limit we get $\varprojlim_m A_m\simeq \hat{A}.$