Most general setting for fine-grained exponential-time complexity classes?
(Just now noticed this question.) There are a lot of questions in the above question. I will try to just address the last few.

Might it be the case that a RAM program can solve general CNF-SAT in exponential time with a base less than 2, but also requiring exponential space, so that when translated to a TM the algorithm runs in exponential time with a base at least 2 and (a slightly-weakened version of) SETH is true in the TM model only?

SETH is generally stated in the word RAM model (in particular, with $O(\log t)$ word size, where the running time is $t$). This is mainly because most (if not all?) algorithmists don't care about or work in the multi-tape Turing machine model, but rather a random-access style model. It is true that in general we do not know how to simulate such time-t RAMs with multi-tape Turing machines super-efficiently, and (more importantly) we do not know how to simulate such RAMs in less than $O(t^2)$ gates with a Boolean circuit. But see the answer to the next question.

Is there a variant of SETH where only subexponential space algorithms are considered in order to make it applicable to more models? Or is there some reason I'm not understanding that no generality is lost without this assumption?

Note there is a very simple time-space tradeoff one can obtain, if SETH is false. Suppose we have a $2^{\beta n}$ time algorithm $A$ for k-SAT on $n$ variables, using $\Theta(2^{\beta n})$ space, where $\beta &lt; 1$. Consider the following algorithm $A_{\alpha}$ parameterized by some parameter $\alpha \in (0,1)$:
On $k$-SAT formula $F$, enumerate all assignments $a$ to the first $n-\alpha n$ variables of $F$, plug $a$ into $F$ and run $A$ on the resulting $\alpha n$-variable formula. Stop and output a SAT assignment if it's found.
Note the above $A_{\alpha}$ runs in $O(2^{n-\alpha n}\cdot 2^{\beta \alpha n}) \leq O(2^{n-(1-\beta)\alpha n})$ time and uses only $\Theta(2^{\beta \alpha n}+\alpha n)$ space.
Corollary: If $k$-SAT has a $2^{\beta n}$ time algorithm, then for every $\epsilon \in (0,\beta)$ there is a $\delta &gt; 0$ such that $k$-SAT can be solved in $O(2^{\epsilon n})$ space and $O(2^{(1-\delta)n})$ time.
Proof: Let $k$-SAT have a $2^{\beta n}$ time algorithm. For any given $\epsilon \in (0,\beta)$, set $\alpha = \epsilon/\beta &lt; 1$. Applying the above reduction, we can solve $k$-SAT in $O(2^{\epsilon n})$ space and $2^{n-(1-\beta)\alpha n}$ time, so $\delta = (1-\beta)\alpha$ suffices. QED
However, note the time-space product of the resulting algorithm is not very favorable...
I do think that the &quot;polynomial-space SETH&quot; (which I believe I have seen in the literature, but am blanking on a reference) that only rules out polynomial-space faster-than-$2^n$ time for $k$-SAT, is a much more plausible hypothesis than SETH itself.