How hard is finding restricted assignment of 3-SAT satisfying $7/8$ of the clauses?
The PCP theorem states that for every $\delta &gt; 0$, it is NP-hard to decide whether a 3CNF is satisfiable or at most $(7/8+\delta)$-satisfiable. Take such a formula $\varphi$ and add to it lots of copies of $(a \lor b \lor c) \land \cdots \land (\lnot a \lor \lnot b \lor \lnot c)$, where $a,b,c$ are new variables (by $\cdots$ I mean eight clauses, exactly seven of which are always satisfied). Suppose these clauses take up $(1-\epsilon)$ of the new 3CNF. This replaces the original $1$ vs. $7/8 + \delta$ to $\epsilon + (1-\epsilon)7/8$ vs. $(7/8+\delta)\epsilon + (1-\epsilon)7/8$, that is $7/8 + \epsilon/8$ vs. $7/8 + \delta\epsilon$. That's not exactly what you wanted since $\delta \neq 0$, but it's pretty close.