I agree with the modifications you performed.

No, this formula count zeroes with their multiplicities.

Let $k \ge 1$. If $p_1,\ldots,p_{2k-1}$ are distinct fixed prime numbers in $[n]$, then the set of all integers which are divisible by at least $k$ prime numbers among $p_1,\ldots,p_{2k-1}$ is coprime-free and looks maximal. Your example corresponds to the case where $k=2$ and $p_1,p_2,p_3$ are $2,3,5$.

No, $\nabla f(x) = \lambda \nabla g(x)$ for some real number $\lambda$ if and only if $x$ belongs to some eigenspace of $u$.

The definition of separability is ambiguous. I guess that the subset $N$ and the sequence $(t_j)$ depend on $J$ and $F$. If yes, one should write « for every J and F, there exists a sequence $(t_j)$ ...»

Yes you are right. I read the assumptions too quickly and I will modify my answer.

What is the bisector of two points?

I am not sure to understand the first question. The symmetry is just that for every vectors $x$ and $y$, $u(x) \cdot y = x \cdot u(y)$, which follows from the definition of $u$. To have $||x||^2=1$, you choose a unit eigenvector for the least eigenvalue. If you take a eigenvector with norm $r$, you minimize $f$ on the sphere $S(0,r)$.

@mattica. You mean that $x_{n_i}$ is the first coordinate of $n_i$?