Sir for S2, can we apply this approach:
We can say that problem L1 can be reduced to problem L2 since if we can solve L2(i.e. if TM for L2 halts, then TM for L1 will also halt), then we can also solve L1. If a string is present in L2, then the strings will surely be present in L1 and so TM for L1 will halt. If a string is not present in L2, then TM for L1 will also halt specifying the strings are not in L1 too. As L2 is recursive, so TM for L2 wont loop forever for any input string and hence TM for L1 wont also loop forever when given the same input strings.Therefore, from reducibility theorem, as L2 is recursive, so L1 is also recursive.