Does there exist a regular octahedron such that... Does there exist a regular octahedron such that it is inscribed in the unit cube and octahedron's vertices belong to $\bf\Large{exactly\; 8}$ edges of the cube?
I think if it exist then octahedron's center can't coincide with cube's center.
 A: Yes. Consider a main diagonal.
Two planes at the same distance from the center, orthogonal to this diagonal cut off triangular pyramids if they are between $\frac16$ and $\frac12$ of the diagonal length away from the center. By continuity (intermediate value theorem), at some suitable distance the six points of intersection make up an octahedron.
A: Theorem: If $C$ is a cube, there exists no inscribed regular octahedron whose vertices belong to exactly eight edges of $C$.
A point $p$ of the $1$-skeleton of $C$ lies on either three edges or one, depending whether $p$ is a vertex or not. 
Let $O$ be an octahedron whose vertices lie on exactly eight edges of $C$. At least one vertex of $O$ must be a vertex of $C$, since $O$ has only six vertices.
$O$ and $C$ cannot have two or more common vertices.
Proof Assume from now on that $C$ is a unit cube. Note that a regular octahedron with unit-length edges has volume $\sqrt{2}/3$. Suppose two vertices $p_1$ and $p_2$ of $O$ are vertices of $C$.
If $p_1$ and $p_2$ are adjacent in $O$ and not adjacent in $C$, then the edges of $O$ have length at least $\sqrt{2}$, so the volume of $O$ exceeds the volume of $C$.
If instead $p_1$ and $p_2$ are adjacent in $C$ (so they lie in exactly five edges of $C$), then the remaining four vertices of $O$, which all lie within a cylinder of radius $1$ about the edge $p_1p_2$, do not touch the $1$-skeleton of $C$.
Finally, "antipodal" vertices $p_1$ and $p_2$ of $O$ cannot both be vertices of $C$; they would have to be antipodal in $C$ as well (since all other pairs of vertices lie in some face of $C$), and a unit cube does not contain three mutually-perpendicular axes of length $\sqrt{3}$ through its center. (There are precisely four segments of length $\sqrt{3}$ in a unit cube, and no two are orthogonal.)
It follows immediately that three or more vertices of $O$ cannot be vertices of $C$.
$O$ and $C$ cannot have exactly one vertex in common.
Proof If $O$ and $C$ have exactly one common vertex and the vertices of $O$ lie in exactly eight edges of $C$, then every vertex of $O$ lies on an edge of $C$. (The common vertex lies in three edges of $C$, and each of the remaining five vertices of $O$ must lie on an edge of $C$.)
If $p$ is a vertex of both $O$ and $C$, then the four vertices of $O$ adjacent to $p$, which are vertices of a square, lie in some sphere centered at $p$. But clearly no sphere centered at $p$ hits exactly four edges of $C$.
This exhausts all possibilities.
A: Yes.  Let the cube be $[0,1]^3$.  Following Hagen von Eitzen's suggestion, the coordinates of the six points are:
$$
(3/4,0,0),\quad (0,3/4,0),\quad (0,0,3/4),\quad (1/4,1,1),\quad (1,1/4,1),\quad (1,1,1/4).
$$
Note that the distance from $(3/4,0,0)$ to $(0,3/4,0)$ is the same as the distance from $(3/4,0,0)$ to $(1,1/4,1)$.
