Answer
(a) $S(2)=10.4ft$, $S(6)=0ft$, $S(8)=3.5ft$, $S(11.75)=91.5ft$
(b) see graph.
(c) $ t=3,9$ for 9:00 a.m. and 3:00 p.m.
(d) $S\to\infty$
Work Step by Step
(a) Since $t=0$ is 6 A.M. 8:00 am is $t=2$ and $S(2)=6|cot\frac{2\pi}{12}|=10.4ft$
Similarly, at noon $t=6$ and $S(6)=6|cot\frac{6\pi}{12}|=0ft$
at 2:00 p.m. $t=8$ and $S(8)=6|cot\frac{8\pi}{12}|=3.5ft$
at 5:45 p.m. $t=8+3+\frac{45}{60}=11.75$ and $S(11.75)=6|cot\frac{11.75\pi}{12}|=91.5ft$
(b) see graph.
(c) Based on the graph, when $S=6, t=3,9$ which correspond to 9:00 a.m. and 3:00 p.m.
(d) When $t\to12, |cot(\pi)|\to\infty, S\to\infty$