The answer to all your questions is no, in general. Simple example:
Here the left picture has one $5$-cycle face and three $3$-cycle faces, while the right picture has two $3$-cyle faces and two $4$-cycle faces. Both outer faces are of different lengths.
But all is not lost. The issue with the above example is that the graph is only $2$-connected. If a planar graph is $3$-connected, then the answer to your first two questions are yes! Very roughly speaking (see Diestel's book for the gory details), when $G$ is a planar $3$-connected graph, then the cycles that are faces (facial cycles) are exactly those induced cycles whose deletion does not disconnect $G$. Since this is a purely graph-theoretical property of the cycles, it implies affirmative answers to your first two questions.
I seem to recall a result stating that you can choose any facial cycle to be the outer face, so I think the answer to your third question is always no unless all faces have the same length, but I can't find a reference, and so I might be misremembering something. In any case, you can find a counterexample pretty easily: Here are two embeddings of a $3$-connected graph, but with different lengths of the outer cycle. Note that they both have two $5$-cycle faces and five $4$-cycle faces however.
(PS sorry about the childish-ly drawn pictures.)