part (iii) is the only tricky part

let A be event that letter to A is placed in correct envelope,
let B be event that letter to B is placed in correct envelope.

method I.
if A is placed in envelope B, then B will definitely end up in an incorrect envelope ,
else if A is placed in envelope C,D or E, then there are 4 envelopes left for B, of which 3 are incorrect envelopes.
P(A'∩B’)
=1/5×1 + 3/5×3/4
=13/20

method 2.
P(A'∩B’)
=P((A∪B)’)
=1-P(A∪B)
=1-[P(A)+P(B)-P(A∩B) ]
=1- (1/5 + 1/5 - 1/5×1/4)
=13/20

note that for part (iii) ...
4/5 × 4/5, or
4/5 x 3/5 , or
1 - 1/5x1/5 .
... all give incorrect answers because A & B are not independent events.