First, the set is $ S=\{9,15,21,25,27,33,35,39,45,49,51,55,57,63,65,69,75\}$. (a) Obviously $ S=\{9,15,21,27,33,39,45,51,57,63,69,75\}\cup \{25,35,45,55,65\}\cup\{49\}$. (b) Since $ 9$ and $ 15$ appear in the set, we need the first progression above for sure. Then we are left with the numbers $ \{25,35,45,49,55,65\}$, the distance between which is $ 10,10,4,6,10$ respectively, so the distance in the progression would need to be $ \gcd(4,6,10)=2$, so the progression should contain all odd numbers, contradiction.