Complexity of specific cases of MAX2SAT
You can always satisfy at least half of clauses: for each variable $x$, find the number of clauses that contain $x$ and the number of clauses that contain $\lnot x$. Select the one which satisfies the most clauses. Remove clauses containing $x$ and $\lnot x$. Repeat for other variables.
Since for each $x$ we satisfy at least half of removed clauses, we satisfy half of the clauses overall.
On the other hand, it's also tight: let $\alpha &gt; \frac 12$ be the fraction of clauses for which we can give an answer. Let $\beta &gt; \frac 12$ be the maximum fraction of clauses we can satisfy in a specific clause. Then we can add clauses so that $\beta$ (for the new clause) becomes arbitrary clause to $\alpha$:

If $\beta &lt; \alpha$, then we can add clauses $(x_i \lor \lnot x_i)$, until $\beta &gt; \alpha$ (since these clauses are always true, $\beta$ increases).
If $\beta &gt; \alpha$, we can add clauses $(x_i)$ and $(\lnot x_i)$, until $\beta &lt; \alpha$ (since exactly half of clauses is true, $\beta$ decreases).

I didn't check, but to get $O(\frac 1m)$ difference (i.e. to find the exact number of clauses), I think it suffices to add $O(m)$ clauses. In other words, if we can solve for some $\alpha &gt; \frac 12$, we can check for any $\beta$ whether $\beta$ fraction of clauses can be satisfied, and therefore we can solve MAX2SAT in polynomial time.