(a) Treat SSSS as one object. Together with P, O, O, E, I, N, there are seven objects, with two identical Os, giving \(7!/2!\) arrangements. If the Os are adjacent, treat OO as another object: \(6!\) arrangements. Thus \[\frac{7!}{2!}-6!=2520-720=\boxed{1800}.\]

(b) Fix P and N at positions 1 and 10. Within the eight remaining positions, the Os must be five positions apart. Their possible original-position pairs are \((2,7),(3,8),(4,9)\). For each pair, arrange four Ss, E and I in the six spaces in \(6!/4!\) ways: \[3\frac{6!}{4!}=\boxed{90}.\]

(c) Identical letters are not distinguished in a selection. The four other letters P, E, I and N are distinct. List the possible counts of S and O:

\[\begin{array}{cc|c|c}\text{Ss}&\text{Os}&\text{other letters}&\text{number of selections}\\\hline2&1&2&\binom42=6\\2&2&1&\binom41=4\\3&1&1&\binom41=4\\3&2&0&1\\4&1&0&1\end{array}\]

The total is \(6+4+4+1+1=\boxed{16}\).