Counter-example: If $J$ is prime then $f^{-1} (J) $ is prime. $f$ need not be unital.
Yes: the zero map would then be a morphism of rings, and the preimage of every prime ideal is then $R$, which is not a prime ideal (by definition)
Yes: the zero map would then be a morphism of rings, and the preimage of every prime ideal is then $R$, which is not a prime ideal (by definition)