Example for first question:
$\Delta^2 u=0$, say on the unit disk, with boundary condition ${\partial\over\partial n}\Delta u=0$, $\Delta u-u=0$. The boundary conditions do not satisfy the Lopatinskii condition (note that the $u$ term is lower order). Nevertheless, it follows from the first boundary condition that $\Delta u$ is constant, and then the second boundary condition implies that $u$ is constant on the boundary. So the kernel is one-dimensional.
Example for second question:
ODE problems like $(x^2y')'-y=0$ with Dirichlet conditions $y(1)=y(-1)=0$.