L1→ Regular, L2→CFL
L3→ REC
L4→ RE
I. ¯L3∪L4 is RE
¯L3∪L4=¯¯¯¯¯¯¯¯¯¯¯¯REC∪RE=REC∪RE
=RE ∪RE=RE
So I is true
II. ¯L2∪L3 is recursive
¯L2∪L3=¯¯¯¯¯¯¯¯¯¯¯¯CFL∪REC=¯¯¯¯¯¯¯¯¯¯¯CSL∪REC
=CSL ∪ REC
=REC ∪ REC=REC
So II is True.
III. L∗1∩L2 is CFL
L∗1∩L2 = (REG)∗∩CFL
=REG∩CFL=CFL
So III is True
IV. L1∪¯L2 is CFL
L1∪¯L2=REG∪¯¯¯¯¯¯¯¯¯¯¯¯CFL=REG∪¯¯¯¯¯¯¯¯¯¯¯CSL
=REG∪CSL=CSL
Since, A CSL need not be a CFL,So IV is False
So only I,II and III are true.