The answer to 2 and 3 is "No".

For (2), observe that a positive answer here would imply all Delaunay simplices to be unimodular (i.e., have volume equal to $1/n!$ times the volume of a fundamental parallelepiped). This holds for $n\le 4$ but starts to fail for $n\ge 5$. See, for example, my paper "Lattice Delone simplices with super-exponential volume" (arXiv, journal) and the references therein.

For (3) I can give an explicit counter-example in dimension three. Consider the lattice generated by $(1,a,a)$, $(a,1,a)$ and $(a,a,1)$, for a very small positive number $a$. When $a=0$ the Delaunay "triangulation" is not a triangulation, but the decomposition of $R^3$ into unit cubes. The effect of making $a\ne 0$ (but small) is that these cubes are refined into simplices, all of which have (by continuity) their center very close to $(1/2,1/2,1/2)$. But it turns out that one of these simplices has the origin and the three basis vectors as vertices, so the center of the circumsphere is outside of it.

(Well, to be precise, the Delaunay "triangulation" of MY lattice is still not a triangulation; in it the unit cube is decomposed into two tetrahedra and one octahedron. But perturbing the coordinates of the basis further, and generically, will produce a true Delaunay triangulation with the same properties).

I am not sure about question 1.

The answer to 2 and 1 is "No".

For (2), observe that a positive answer here would imply all Delaunay simplices to be unimodular (i.e., have volume equal to $1/n!$ times the volume of a fundamental parallelepiped). This holds for $n\le 4$ but starts to fail for $n\ge 5$. See, for example, my paper "Lattice Delone simplices with super-exponential volume" (arXiv, journal) and the references therein.

For (1) I can give an explicit counter-example in dimension three. Consider the lattice generated by $(1,a,a)$, $(a,1,a)$ and $(a,a,1)$, for a very small positive number $a$. When $a=0$ the Delaunay "triangulation" is not a triangulation, but the decomposition of $R^3$ into unit cubes. The effect of making $a\ne 0$ (but small) is that these cubes are refined into simplices, all of which have (by continuity) their center very close to $(1/2,1/2,1/2)$. But it turns out that one of these simplices has the origin and the three basis vectors as vertices, so the center of the circumsphere is outside of it.

(Well, to be precise, the Delaunay "triangulation" of MY lattice is still not a triangulation; in it the unit cube is decomposed into two tetrahedra and one octahedron. But perturbing the coordinates of the basis further, and generically, will produce a true Delaunay triangulation with the same properties).

I am not sure about question (3), but would expect a negative answer as well.

The answer to 2 is "No". Observe that a positive answer here would imply all Delaunay simplices to be unimodular (i.e., have volume equal to $1/n!$ times the volume of a fundamental parallelepiped). This holds for $n\le 4$ but starts to fail for $n\ge 5$. See, for example, my paper "Lattice Delone simplices with super-exponential volume" (arXiv, journal) and the references therein.

I am not sure about the other two questions; I would be very surprised if 3 is true, and have no clear opinion on 1.

The answer to 2 and 3 is "No".

For (2), observe that a positive answer here would imply all Delaunay simplices to be unimodular (i.e., have volume equal to $1/n!$ times the volume of a fundamental parallelepiped). This holds for $n\le 4$ but starts to fail for $n\ge 5$. See, for example, my paper "Lattice Delone simplices with super-exponential volume" (arXiv, journal) and the references therein.

For (3) I can give an explicit counter-example in dimension three. Consider the lattice generated by $(1,a,a)$, $(a,1,a)$ and $(a,a,1)$, for a very small positive number $a$. When $a=0$ the Delaunay "triangulation" is not a triangulation, but the decomposition of $R^3$ into unit cubes. The effect of making $a\ne 0$ (but small) is that these cubes are refined into simplices, all of which have (by continuity) their center very close to $(1/2,1/2,1/2)$. But it turns out that one of these simplices has the origin and the three basis vectors as vertices, so the center of the circumsphere is outside of it.

(Well, to be precise, the Delaunay "triangulation" of MY lattice is still not a triangulation; in it the unit cube is decomposed into two tetrahedra and one octahedron. But perturbing the coordinates of the basis further, and generically, will produce a true Delaunay triangulation with the same properties).

I am not sure about question 1.