Among finite lattices, no. A finite lattice (with more than 2 elements) has atoms. Let $a$ be an atom. Condition 2 says there must be an incomparable $b$ that meets $a$ above bottom, which is clearly impossible.
Among infinite lattices, yes. Consider the following lattice, consisting of two slanted infinite ladders, and augmented bottom and top elements. If $a$ is in the left ladder (say), it has infinitely many complements in the right ladder, satisfying condition 1. Also, there is an element $b$ in the same ladder (pictured), which is incomparable with $a$, joins below $1$, and meets above $0$, satisfying condition 2.
