Conditions for tractability of 3SAT-Satisfiability
Yes. If $ 0&lt; \epsilon &lt;1$ is a constant (or $1/\textit{polylog}(n)$), and you are promised that at least $ \epsilon 2^n $ of all possible assignments are satisfying the input 3CNFs, then you can find such an assignment in deterministic polynomial-time.

The algorithms is not difficult: 

Claim: Under the promise stated, there must exist a constant size set $ S $ of variables that hits all clauses in the 3CNF, in the sense that every 3-clause must contain a variable from $ S $. 

Proof of claim (sketch): Otherwise, there must exist a large enough family of 3-clauses from the 3CNF, in which each variable occurs only once. But this family, when sufficiently large, has already less than $ \epsilon $ fraction of satisfying assignments. QED

Thus, you can run over all possible (constant number) of assignments to $ S $. Under every fixed assignment to $ S $, the 3CNF becomes a 2CNF, by the assumption that $ S $ hits the original 3CNF. Now, you can use the known polytime deterministic algorithm for finding a satisfying assignment for 2CNF formulas.
Overall, you get a polynomial time upper bound.

The algorithm for 2SAT is I think already in S. Cook famous 1971 paper.

The algorithm for 3CNFs is from:
L. Trevisan
A Note on Deterministic Approximate Counting for k-DNF
In Proc. of APPROX-RANDOM, Springer-Verlag, page 417-426, 2004

The original paper showing the result for 3CNF is:
E. Hirsch, A fast deterministic algorithm for formulas that have many satisfying assignments, Journal of the IGPL, 6(1):59-71, 1998