a) Let the sequence $(a_n)$, $n \in \mathbb{N}$, be lacunar. Then there exists a number $q > 1$ such that
$$a_{n+1} \ge q a_n \quad \forall n \in \mathbb{N}. \quad (1)$$
In particular, any lacunar sequence is increasing. From (1) it follows that any 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 interval, then $\frac{a_{n+1}}{a_n} < \frac{qx}{x} = q$, contrary to (1). Therefore, any lacunar sequence is solitary.

b) Consider the lacunar sequence $a_n = 2^n$, $n \in \mathbb{N}$. This sequence is solitary as was shown in item a): every interval $(x, 2x)$, where $x > 0$, contains at most one term of this sequence. We construct a new sequence $(x_n)$, $n \in \mathbb{N}$, as $x_{2n-1} = a_{2n}$ and $x_{2n} = a_{2n-1}$ for any $n \in \mathbb{N}$. This sequence $(x_n)$ is solitary because the set of its terms and the set of the terms of the sequence $(a_n)$ coincide. But the sequence $(x_n)$ is not lacunar since this sequence is not increasing.