Consequences of $NP=coNP=BPP=RP$
Unless I'm missing something obvious (it is getting late here), the second isn't possible - $BPP = coBPP$, so $NP=BPP \rightarrow NP = coBPP \rightarrow NP = coNP$.

In the first case, some of the consequences would be:


"Feasible" algorithms for $NP$-complete problems, for at least the randomised version of feasible.
$PH \subseteq NP$
Unless $P = NP$, adding randomness adds computational power.
Strong pseudorandom number generators do not exist (nor do one way functions). (we would need $BPP = P$ for them to exist, but that would imply $P=NP$, in which case they don't).