a) Let the sequence $(a_n)$, $n \in \mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that
$$a_{n+1} \ge q a_n \quad \forall n \in \mathbb{N}. \qquad (1)$$
In particular, any lacunar sequence is increasing. From (1) it follows that any half-interval $(x, qx]$ contains at most one term of this sequence. Indeed, if we assume that $a_n$ and $a_{n+1}$ belong to this half-interval, then $\frac{a_{n+1}}{a_n} < \frac{qx}{x} = q$, contrary to (1).

Consider any positive integer $k$ such that $q^k > 2$ (e.g., $k = [\log_2 2] + 1$, where $[y]$ stands for the integer part of $y$). It is evident that
$$(x, 2x) \subset \bigcup_{i=1}^{k} (q^{i-1}x, q^i x]. \qquad (2)$$
Since at most one term of $(a_n)$ belongs to the half-interval $(q^{i-1}x, q^i x]$, from inclusion (2) it follows that at most $k$ terms of $(a_n)$ belong to the interval $(x, 2x)$.

b) We define the sequence $(a_n)$, $n \in \mathbb{N}$, as
$$a_{2n-1} = 2^n, \quad a_{2n} = 2^n \left(1 + \frac{1}{n+1}\right), \quad n \in \mathbb{N}. \qquad (3)$$
This sequence is increasing since from (3) it follows that $a_{2n-1} < a_{2n} < a_{2n+1}$ for all $n \in \mathbb{N}$. Moreover, this sequence is rare since at most one number of the form $2^m$, $m \in \mathbb{N}$, belongs to any interval $(x, 2x)$, and so at most three terms of this sequence belong to this interval.

On the other hand, we have
$$\frac{a_{2n}}{a_{2n-1}} = 1 + \frac{1}{n+1}, \quad n \in \mathbb{N},$$
but the numbers $1 + \frac{1}{n+1}$ may be arbitrary close to $1$ (for $n$ large enough), hence, the sequence $(a_n)$, $n \in \mathbb{N}$, is not lacunar.