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The answer to question (2) is

• no for $n=6$,
• yes for $n=7$.

For $n=6$, take, for example, the following six points as vertices of a straight-line (stick) embedding of $K_6$:

$A = (-2,-2,1), B= (2,-2,0), C= (0,2,0), D= (-1,-1,0), E= (1,-1,1), F= (0,1,2)$

The projection onto the $xy$ plane has crossing number $3$.

Moreover, the crossings are between disjoint pairs of edges. Therefore, since every nontrivial knot has at least three crossings, there is at most one possible cycle that could form form a nontrivial knot; that is, the cycle $AECDBF$ formed by the six edges participating in the crossings. But by the above-below relations at the crossings, this cycle clearly forms an unknot.

For $n=7$, Conway and Gordon proved that every embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a nontrivial knot, using the parity of the sum of the quadratic terms of the Conway polynomials of the Hamiltonian cycles as an invariant.

Edit: See also J. L. Ramirez Alfonsin, Spatial Graphs and Oriented Matroids: the Trefoil, Discrete and Computational Geometry 22:149--158 (1999) for the following stronger result:

Every stick embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a (left-handed or a right-handed) trefoil.

1

The answer to question (2) is

• no for $n=6$,
• yes for $n=7$.

For $n=6$, take, for example, the following six points as vertices of a straight-line (stick) embedding of $K_6$:

$A = (-2,-2,1), B= (2,-2,0), C= (0,2,0), D= (-1,-1,0), E= (1,-1,1), F= (0,1,2)$

The projection onto the $xy$ plane has crossing number $3$.

Moreover, the crossings are between disjoint pairs of edges. Therefore, since every nontrivial knot has at least three crossings, there is at most one possible cycle that could form form a nontrivial knot; that is, the cycle $AECDBF$ formed by the six edges participating in the crossings. But by the above-below relations at the crossings, this cycle clearly forms an unknot.

For $n=7$, Conway and Gordon proved that every embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a nontrivial knot, using the parity of the sum of the quadratic terms of the Conway polynomials of the Hamiltonian cycles as an invariant.