for $ a$: if we find $ k$ consecutive numbers which are not the power of any integer,we are obviously done(because each of these numbers can be written in the form $ ab$ where $ \gcd(a,b)=1$).that is easy(see E27,PEN).

for $ b$: i hope that by saying "cardinality of a set" u mean the number of its elements(sorry for my poor english ), if so, then u can easily see that $ f(n+1)>f(n)$ increases iff $ n+1$ is the power of a prime.so if we want to find such sets like $ A$ s.t. $ A=\{n,n+1,...,n+t\}$ there wil be 2 cases: $ n$ is even $ \implies n=2^k,n+1=p^{\alpha},n+2=2^{\beta},...$.this case is trivial. $ n$ is odd $ \implies n=p^{\alpha},n+1=2^k,n+2=q^{\beta},...$.in this case u can easily see that $ k$ is a prime which implies that $ q=3$.the rest is easy,too.

Part a)Choose $X=N!+1$,then $N!+1+j=x+j=(j+1) \cdot (\frac {N!}{j+1}+1) $ for all $j <n$,and it is clear that this construction works,since both of the terms are less than $N $($(j+1)^2|N! $ for all $n>j$ which is clearly possible)