2 SAT variants complexity class
b. is completely trivial - just set each variable to FALSE.

d. trivially reduces to c. ​ ​ ​ c. reduces to d. as follows:

For each double-negative clause, remove the negatives. ​ Replace each

single-negative clause with 2 clauses, each of which has one of the original clause's variables

and a new variable that's shared with the other of the 2 new clauses but otherwise unused.

That works because:

for 2-variable clauses, NAE is just ≠

for boolean variables, ​ ​ ​ x ​ NAE ​ not y ​ ​ ​ simplifies to ​ x = y

for boolean variables, [there exists a y such that ​ ​ ​ x ≠ y ​ and ​ y ≠ z ] ​ if and only if ​ x = z

By ​ ​ ​ variables &lt;-> vertices ​ and ​ clauses &lt;-> edges ​ , ​ ​ ​ c. is equivalent to bipartiteness.

Bipartiteness is logspace-complete. ​ (Thus, c. and d. are logspace-complete.)

By throwing in k extra variables to handle "at most" vs. "exactly", a. is in W[P].

By ​ ​ ​ vertices -> variables ​ and ​ edges -> clauses ​ , ​ ​ ​ vertex cover reduces to ​ a. .

(Thus, for unrestricted k, a. is NP-complete.) ​ However, a. is fixed-parameter tractable.

Proof:
By replacing each clause with k+1 copies of itself and then [for each variable, putting in a

clause whose literals are both the negation of that variable], a. reduces to Almost 2-SAT.

By part 2.1 of that paper, that paper handles the case of clauses whose 2 literals are the same.

Theorem 7 of that paper explicitly covers "possible repeated occurrences of clauses".

I have no clue regarding possible kernalizability of a. or

$\big[$the space requirements of algorithms for a. whose runtimes aren't ​ max$\hspace{-0.03 in}\big(\hspace{-0.04 in}$nΩ(k),$\hspace{.02 in}$2Ω(n)$\hspace{-0.03 in}\big)\big]$.