I do for a,b similar, maybe i do not true, check for me Let $a_1=p_{1}$ Let $a_{m}=lcm(a_{1},a_{2},..,a_{d(m)+e}).p_{m+2}^{k_{m}}$ Here: e=-1 for a and =1 for b $k_{m}$ is enough large number such that $a_{m}>a_{r}$ with k>r If $a_{k}\mid a_{m}$ for $m> k>d(m)+e$ then $p_{k+2} \mid a_{m}$ that is not true. Then that sequence $a_{m}$ exist.

Exactly that

a) Since $\gcd(2^m-1,2^n-1)=2^{\gcd(m,n)}-1$, the sequence $a_1=2$ and $a_n=2^n-1$ for $n\geq2$ has this property. b) Yes. Let $a_1,\dots,a_N$ be a sequence of primes, and for $k>N$, let $a_k=p_k\cdot\operatorname{lcm}(a_1,\dots,a_{d(k)})$, where $p_k>a_{k-1}$ is prime. Clearly now $a_1,\dots,a_{d(m)},a_m$ are the only terms dividing $a_m$.

what is d(m) ?