Is there a planar graph that (almost) all its vertices have degree 6? Is it true that for any $N_0\in\mathbb N$ there exists a planar graph $G=(V,E)$ on (at least) $N_0$ vertices such that at least $$|V|(1-o(1))$$ vertices has degree 6?
It is easy to show that no 6-regular planar graph exists, but is there an "almost-6-regular" planar graph?
 A: Take a regular icosahedron whose edges are unit length.
For any $n > 1$, subdivide each of its face into $n^2$ equilateral triangle with edge length $1/n$. You obtain a polyhedron with $10 n^2 + 2$ vertices, $30n^2$ edges and  $20 n^2$ faces.
Using stereographic projection from the center of one of its face,
it is easy to see the vertices and edges of this figure is a planar graph.
Aside from the original $12$ vertices, the other $10(n^2 - 1)$ vertices have degree $6$.
A: Consider a hexagonal tiling of the plane, and subdivide each hexagon into six equilateral triangles.  Now all the vertices have degree six, albeit with an infinite number of vertices.
If you are willing to settle for merely a high proportion of vertices with degree six, consider truncating the tiling at a large radius $R$ from the origin.  The vertices inside this radius will still have degree six, and only the boundary vertices will have fewer than six neighbors.  But the count of vertices inside is proportional to the area covered, so $O(R^2)$ while the count of vertices on the boundary is proportional to the perimeter, and that is easily seen to be $O(R)$.
Thus as $R$ increases without limit, the proportion of degree six vertices tends to 1.

It might be less "hand wavy" to consider a single regular hexagon, subdivided into six equilateral triangle, then subjected to successive refinements in which every triangle is subdivided into four equal sub-triangles (with the accompanying bisection of sides).  The number of triangles grows by a factor of 4 with each refinement, which the number of vertices on the perimeter only doubles with each step.
