i) Candidate keys are C and AB .
D -> E is not satisfy BCNF property .
R1(DE) and R2(ABCD) in BCNF
Dependencies in R1 are {D -> E } . In R2 are {AB->C, C->D , C->AB}
ii) Candidate keys are AB and BC .
C -> D and C-> E are in 1 NF
Decompose into R1(ACDE) and R2(BC) .
Dependencies are {C -> A , C-> D , D -> E}
AB -> C dependency is not preserved
Hence Answer is a)