5 cards from 30 can be selected in 30C5 ways
here every order is considered as distinct, hence total number of selections are 30C5*5!

i)
two cards 5 and 28 to be included, this can be done in 1 way and remining 3 cards from 28 can be drawn in 28C3 ways
and 5! are the arrangements
Required probability = 28C3*5!/30C5*5! = 0.0230

ii)
remaining 3 to be selected from 28 cards in 28C3 ways and can be arranged in 3! ways

Required probability = 28C3*3!/30C5*5! = 0.0011

iii)
If we dont want to select 15, there are 29 cards available to select 5 cards

Required probability = 29C5/30C5 = 0.8333