(a) Using name initials for the people, we can see that if I and N never sit together, there can only be one or two spaces between them. The other people have 6 ways they can be arranged. So if I and N are seated with one person between them there are 6 arrangements for the other 3 people, and if I and N are seated with two people between them there are another 6 arrangements for the others. Remember also that they are sitting at a round table, so if I and N have one person between them clockwise, they automatically have two people between them anticlockwise. This makes 12 different ways altogether so that I and N are not sitting together:
IJNPS, IJNSP, IPNJS, IPNSJ, ISNPJ, ISNJP, IJPNS, IJSNP, IPJNS, IPSNJ, ISPNJ, ISJNP all fix the position of I as the starting point in the circle. This covers all cyclic arrangements. Note that there are 8 occurrences of I next to each of J, P and S. Similarly for N.
(b) If IN are together and there are 6 arrangements for the others, IN can be sandwiched in 3 places between the 3 other people, making 18 arrangements in all, or so it appears, because the cyclic arrangement gives us duplicates, which I've struck through:
JINPS, JPINS, JPSIN, JINSP, JSINP, JSPIN, PINJS, PJINS, PJSIN,
PINSJ, PSINJ, PSJIN, SINPJ, SPINJ, SPJIN, SINJP, SJINP, SJPIN.
We can therefore remove the struck through arrangements as duplicates, leaving 6 so far.
However, we can reverse the seating of IN to NI:
JNIPS, JPNIS, JPSNI, JNISP, JSNIP, JSPNI.
This gives 12 cyclic arrangements for I and N sitting together. Note that there are 4 occurrences of I and N next to each of J, P and S.