Traverse triangles in Triangulation of 2-sphere without repetition (Hamiltonian Path) Let $T_0, \dots, T_n$ be a collection of spherical triangles that form a triangulation of the $2$-sphere. 
Can we enumerate the triangles in such a manner that $T_{i-1}$ and $T_{i}$ have exactly one common edge for all $1 \leq i \leq n$? Can we enumerate them such that $T_n$ and $T_0$ have one common edge?
This question can be seen through the lense of graph theory: let $V$ be a graph whose vertices are the triangles, and let $E$ be the set of edges (quite literally). The question is whether the graph G=(V,E) has a Hamiltonian path or a Hamiltonian tour.
 A: Nope!
Finding a Hamiltonian cycle is almost equivalent to Tait's Hamiltonian Graph Conjecture
that every cubic polyhedral (i.e. 3-connected planar) graph is Hamiltonian. That is because every cubic polyhedral graph gives you a convex polyhedron with all 3-valent vertices, whose dual is a polyhedron with all triangular faces, which induces a triangulation of the 2-sphere.
The counterexamples to Tait's conjecture thus give you triangulations of the 2-sphere with no Hamiltonian cycle among the triangles.
For Hamiltonian paths, the question is whether every 3-connected planar cubic graph is traceable. I didn't find anyone discussing merely untraceable graphs, but there are several papers looking more specifically for hypotraceable graphs (which have no Hamiltonian path, but removing any vertex leaves a graph with a Hamiltonian path).
Carsten Thomassen's 1981 paper "Planar Cubic Hypohamiltonian and 
Hypotraceable Graphs" (DOI 10.1016/0095-8956(81)90089-7) has

THEOREM 3.1. There exist infinitely many planar cubic hypohamiltonian 
  graphs and infinitely many planar cubic hypotraceable graphs.

Investigating the relevant construction from his 1976 paper "Planar and infinite hypohamiltonian and hypotraceable graphs" (DOI 10.1016/0012-365X(76)90071-6),
we find that it yields 3-connected graphs, so these are polyhedral and give rise to untraceable 2-sphere triangulations.
The smallest of these hypotraceable graphs has 460 vertices.
"On cubic planar hypohamiltonian
and hypotraceable graphs" (M. Araya, G. Wiener, 2011)
finds smaller examples, in

Corollary 3.2. There exists a cubic planar hypotraceable graph on 340
  vertices and on $n$ vertices for every even number  $n ≥ 356$.

