The problem $3$-SAT and $2$-SAT are
both in $P$
both $NP$ complete
$NP$-complete and in $P$ respectively
undecidable and $NP$ complete respectively
Option is $C$.
Why 2-SAT problem is solvable in polynomial time by deterministic machine ? ..It is beautifully explained by Prof. Shai Simonson here :-
$3$-SAT is NP complete
$2$-SAT is P
Option (c) NP-complete and in P, respectively is the correct answer.
3-SAT and 2-SAT are special cases of k-satisfiability (k-SAT) or simply satisfiability (SAT) when each clause contains exactly k = 3 and k = 2 literals respectively.
2-SAT is P while 3-SAT is NP-Complete.