Consider a finitely supported random walk on a discrete group G such that the support generates $G$ as a semigroup. The Martin kernels are then non-negative functions on the product $G \times M$ where $M$ denotes the Martin boundary of the random walk. Does anyone know an example with $G$ amenable such that there does not exist a point m in the Martin boundary for which the function $K( . , m)$ is constant $1$? And does anyone know an example of a non-amenable group for which there is an element $m$ in the Martin boundary such that the corresponding Martin kernel $K( . ,m)$ is constant $1$?
The Martin boundary of any amenable group determined by a symmetric probability measure always contains the function 1 - this is a result of Northshield. I am not aware of any answers to your other questions which seem to be quite interesting.
EDIT This is an observation (too long for a comment) concerning the answer of M. Dus below. If the minimal Martin boundary (i.e., the subset of the Martin boundary which consists of minimal harmonic functions) is closed, then the group action on the minimal Martin boundary is topologically amenable. The groups that admit topologically amenable actions are called amenable at infinity. Therefore, for any group which is not amenable at infinity the minimal Martin boundary is not closed (in particular, does not coincide with the whole Martin boundary) for any random walk on the group. See Theorem 6.5 and Section 6.E in this paper by Kaimanovich.
This is a partial answer to your second question. If the function 1 is minimal, then every bounded harmonic function is constant. To reformulate, if there exists a point $m$ in the minimal Martin boundary such that $K(\cdot, m)$ is constant 1,then the Poisson boundary is trivial, which cannot happen if the group is non-amenable. So you have to look at non-minimal points.
EDIT (taking into account the comments of R W)
However, for finitely supported measures, in several classes of groups (such as hyperbolic groups and relatively hyperbolic groups with respect to virtually abelian subgroups), the Martin boundary is minimal. So in particular, for such examples of groups, the answer is no. You thus have to look for groups such that the minimal boundary is not the whole Martin boundary (such as the examples given by R W).