For part c) also check if your decomposition is lossless join and dependency preserving.


1a) 

F  = { C->D,C->A,B->C }
C+ = { A,C,D}
B+ = { A,B,C,D}

B is a key.

1b)

not in BCNF, because of C->AD
not in 3NF, because of  C->AD

1c) 

One possible answer:

S1 = {ACD}, S2 = {BC} 

S1 intersect S2 = {C} and C -> S1, hence lossless-join

dependency preserving because F_S1 U F_S2 = F

2a)

F = {B->C, D->A}

BD+ = {ABCD}

BD is a key.

2b)

not in BCNF because of B->C
not in 3NF  because of B->C

2c)

S1 = {BC}, S2 = {AD}, S3 = {BD}

Chase test confirms lossless join
Dependency preserving because ( {B->C} U {D->A} ) + = F+

3a) 
F = { ABC->D, D->A}

key: ABC

3b)

not in BCNF because of D->A
in 3NF, since A is part of a key 

3c) 

S1={AD}, S2={BCD}

lossless join because intersection is a key to S1
not dependency preserving

4a)

key: A

4b)

not in BCNF because of BC->D
not in 3NF

4c)

S1={BCD}, S2={ABC}

lossless join since intersection is key for S1
dependency preserving.

5a)

AB -> CD hence AB is key

C -> A
BC -> AB (aug B)  hence BC is also key

D->B
AD -> AB (aug A) hence AD is also key

C->A
CD->AD (aug D) hence CD is also key

5b)

not in BCNF because LHS of C->A is not a superkey
in 3NF because RHS of C->A and  D ->B are part of keys