Is there an adaptive version of LWE assumption with respect to some potentially non-uniform secret distribution? There is a version of LWE assumption as follow.

Assume that there is a positive number $n$, an integer $q = q(n) \geq 2$, an error distribution $\chi = \chi_{n}$, a vector $\mathrm{\mathbf{s}} \gets \mathbb{Z}_{q}^{n}$ and for every efficient algorithm $\mathcal{A}$,
$$\Pr \left[ A \gets \mathbb{Z}_{q}^{m \times n}, \mathrm{\mathbf{e}} \gets \chi^m,~  \mathrm{\mathbf{b}} \gets A \cdot \mathrm{\mathbf{s}} + \mathrm{\mathbf{e}}: \mathcal{A}(A, \mathrm{\mathbf{b}}) = s \right] \leq \mathrm{nelg}(n)$$

Q1: Why does it say $m = O(n \log q)$ in some schemes? I do not find some notes about the range of $m$. I think $m = \mathrm{poly}(n \log q)$ is OK.

Q2: Is there an adaptive version? Is it still hard to solve?

For example, Let oracle $\mathcal{O}_{\mathrm{\mathbf{s}}}$ satisfy that  $\mathcal{O}_{\mathrm{\mathbf{s}}}(\mathrm{\mathbf{a}}) = (\mathrm{\mathbf{a}},  \langle \mathrm{\mathbf{a}}, \mathrm{\mathbf{s}} \rangle + e)$, where $e \gets \chi$.
And for every efficient algorithm $\mathcal{A}$,
$$\Pr \left[ \mathcal{A}^{\mathcal{O}_{\mathrm{\mathbf{s}}}}(1^n) = s \right] \leq \mathrm{nelg}(n)$$
The algorithm $\mathcal{A}$ is allowed to ask the access of $\mathcal{O}_{\mathrm{\mathbf{s}}}$ for at most $m$ times.

Q3: If there are $l$ secret vectors and $S = [\mathrm{\mathbf{s}}_{1}, \mathrm{\mathbf{s}}_{2}, \ldots, \mathrm{\mathbf{s}}_{l}]$.
Is there a version of LWE with respect to some potentially non-uniform secret distribution? (e.g., $S$ belongs to the general linear group $GL_{n}(q)$ or the orthogonal group $S \in O_{n}(q)$)

It means that for every efficient algorithm $\mathcal{A}$,
$$\Pr \left[ \mathcal{A}^{
\mathcal{O}_{\mathrm{\mathbf{s}}_{1}}, 
\mathcal{O}_{\mathrm{\mathbf{s}}_{2}}, 
\ldots, 
\mathcal{O}_{\mathrm{\mathbf{s}}_{l}}
} (1^{nl}) = S \right] \leq \mathrm{nelg}(nl)$$
For Q1, $m = O(n\log q)$ is the size required for the leftover hash lemma to kick in and $Ax$ to be statistically close to uniform (I believe). See questions like this one.

For Q2, the answer is yes. The hardness of LWE is essentially independent of the dimension $m$, as you can generate new LWE samples from a fixed collection of samples with only a mild loss in the error term. See proposition 2.1 of Regev's survey (and the "other implications" discussion after the proof).