Gap problems clarifation
You are confusing NP and NP-hard. The fact that a problem is (say) in L doesn't preclude it being in NP, in fact, quite the contrary: any problem which is in L is a fortiori in NP, i.e. L$\subseteq$NP. On the other hand, it is not likely to be NP-hard (and assuming that P$\neq$NP, if it is in L then it cannot be NP-hard).

Another mistake you're making is that a problem which is approximable can't be NP-hard (or rather, its decision version can't be NP-hard). Considering MAX-3SAT, there is a trivial 7/8-approximation algorithm (pick a random assignment; this can be derandomized), yet the decision version 3SAT is NP-complete.

You're right that GAP-A$(a,b)$ being easy doesn't imply that GAP-A$(a/2,b/2)$ is easy. One example (replacing $1/2$ with an arbitrary $c$) is MAX-2SAT. GAP-2SAT$(a,1)$ is easy for any $a&lt;1$, yet there exist some $b,c$ for which GAP-2SAT$(b,c)$ is NP-hard. (Unfortunately, $c=1/2$ doesn't work, since every 2SAT instance can be $1/2$-satisfied.)