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This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

But if there is a $(3,9)$ Fisk triangulation of the torus, then there is also an "odd" Fisk triangulation of an orientable surface of any genus. Instead of cutting along edges, cut the "odd Fisk" torus and a $6$-regular torus along two-edge paths (for the "odd Fisk" torus take a path between the exceptional vertices. The cut open tori can be glued together so that the odd vertices remain odd and adjacent, and all other vertices keep degree $6$. This gluing will not create a double edge as the previous construction.

EDIT: On the other hand, I am starting to doubt the existence of a $(3,9)$-Fisk triangulation of the torus. Maybe one can prove by geometric methods that in the canonical Euclidean cone-metric there are several shortest geodesics between the exceptional points. This would mean, there must be multiple edges in the skeleton of the triangulation.

This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

But if there is a $(3,9)$ Fisk triangulation of the torus, then there is also a $6$-regular Fisk triangulation of an orientable surface of any genus. Instead of cutting along edges, cut the $(3,9)$-torus and a $6$-regular torus along two-edge paths (for the $(3,9)$-torus take a path between the exceptional vertices. The cut open tori can be glued together so that the odd vertices remain odd and adjacent, and all other vertices keep degree $6$. This gluing will not create a double edge as the previous construction.

EDIT: On the other hand, I am starting to doubt the existence of a $(3,9)$-triangulation of the torus. Maybe one can prove by geometric methods that in the canonical Euclidean cone-metric there are several shortest geodesics between the exceptional points. This would mean, there must be multiple edges in the skeleton of the triangulation.

This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

If one wants to have triangulations, it suffices to find a $(3,9)$ Fisk triangulation of the torus. Then one cuts it open along a two-edge path joining the exceptional vertices and glues (in a correct way) to a torus cut open along two edges of a triangle. This will not create a double edge as the previous construction, but only a $4$-cycle.

This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

But if there is a $(3,9)$ Fisk triangulation of the torus, then there is also an "odd" Fisk triangulation of an orientable surface of any genus. Instead of cutting along edges, cut the "odd Fisk" torus and a $6$-regular torus along two-edge paths (for the "odd Fisk" torus take a path between the exceptional vertices. The cut open tori can be glued together so that the odd vertices remain odd and adjacent, and all other vertices keep degree $6$. This gluing will not create a double edge as the previous construction.

EDIT: On the other hand, I am starting to doubt the existence of a $(3,9)$-Fisk triangulation of the torus. Maybe one can prove by geometric methods that in the canonical Euclidean cone-metric there are several shortest geodesics between the exceptional points. This would mean, there must be multiple edges in the skeleton of the triangulation.

This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "triangulation" (which is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$, and I strongly believe there is a simplicial complex with this property.

For the surfaces of higher genus I would guess that one can have all vertex degrees $6$ except two odd neighbors. This guess is based only on the fact that the combinatorial (coloring monodromy) and the geometric (holonomy of a Euclidean cone-metric) obstructions don't forbid this.

This is only a partial answer.

First, a sphere does not admit Fisk triangulations. Fisk has proved it by using the 4-color theorem, but there is a simpler proof using Fisk's method of coloring monodromy. Try to color the vertices in three colors. There will be no contradiction when you go around an even vertex; but the colors will be permuted when you go around an odd vertex. For one vertex this will be the $(12)$ transposition, for the other vertex $(23)$ transposition. Thus, the fundamental group of the sphere minus two points will be mapped epimorphically onto $S_3$, which is a contradiction.

The torus does not allow triangulations with vertex degrees $5, 7, 6, \ldots, 6$ (even if you don't require the $5$- and the $7$-vertex to be neighbors). This is a result by Jendrol and Jukovic, reproved recently by a geometric method. There is however a "pseudotriangulation" (in the sense that it is not a simplicial complex) with two vertices, one of degree $3$, the other of degree $9$. If you cut it open along an edge and glue to a $6$-regular torus cut open along an edge, then you get a surface of genus two with vertices of degrees $9$ and $15$. This can be iterated. (Again, the result is a pseudotriangulation.)

If one wants to have triangulations, it suffices to find a $(3,9)$ Fisk triangulation of the torus. Then one cuts it open along a two-edge path joining the exceptional vertices and glues (in a correct way) to a torus cut open along two edges of a triangle. This will not create a double edge as the previous construction, but only a $4$-cycle.