a.In R1 and R2 the common attribute is M,M+={MO}.We see that M is key for one of the decomposed relation R2,hence the decomposition is lossless.

b,

R1 R2

P->L,L->MN M->O

In R1 FD s covered are P →L and L →MN,in R2 M→O,.

We see that NO →P cannot be directly covered by any of the FD s in the decomposed relation.So this demands finding the additional FD s that can be implied in R1(using P →L and L →M,L->N),as well as R2(using M->O).All these FD s fail to cover NO->P ,as no way we can obtain P in the RHS.

So NO->P is not covered.

c)So till now we get.

R1 R2

P->L,L->MN M->O

R2 being a binary relation is in BCNF,M is SK and is LHS.

In R1 P is the key,

P+={PLMN}.

P is SK LHS and non trivial dependency.So ok.

L is non prime as well as MN is also non prime.So Transitive dependency is present.So not 3 NF.

Key is single attributed ,so no question of any prime attribute to be present.Hence 2 NF.

Hence highest Normal form in R1 is 2NF.

Colored portion is edited.