Complexity of 3SAT variants
3SAT_1 is easy, and a variant of 3SAT_2 is NP-complete. My guess is that 3SAT_2 is also NP-complete. Update: my guess is proved below.

Let $C = x\lor y\lor z$ be a clause. Then $\lnot C = \lnot x \land \lnot y \land \lnot z$. An assignment makes every clause of $\phi$ false if it makes all their negations true. The conjunction of the negation of all clauses is just a big conjunction. A conjunction is satisfiable if and only if it does not contain a variable and its negation. So 3SAT_1 is in P (in fact it's in AC$^0$).

The variant of 3SAT_2 asks whether $\phi$ has an assignment that makes exactly $8/9$ of the clauses true. This is clearly in NP. To reduce 3SAT to this variant, take the formula $\phi$, and for each clause $C$ in $\phi$ add the eight clauses
$$(x_C \lor y_C \lor z_C) \land \cdots \land (\lnot x_C \lor \lnot y_C \lor \lnot z_C).$$
In any assignment, exactly seven of these are correct. In total we have $9|\phi|$ clauses. Of the $8|\phi|$ that we added, exactly $7|\phi|$ are always correct. Hence $\phi$ is satisfiable if and only if in the new formula one can satisfy exactly $8|\phi|$ constraints, which are $8/9$ of its clauses.

Update: Here is a reduction from 3SAT to 3SAT_2 itself. Given a formula $\phi$, consider two cases. The first case is when we can falsify all clauses in $\phi$ at once. As shown above, in this case each variable appears only positively or only negatively, and so setting it appropriately, the formula is satisfiable. Otherwise, add $|\phi|$ clauses of the form $x \lor y \lor z$, where $x,y,z$ are new variables. In any assignment either all of these are satisfied, or all are not satisfied. Since the original clauses cannot be all falsified at once (by assumption), the new formula can be half-satisfied if and only if the original one is satisfiable.