= Solution
{c}
For the first <sequence>, $|a_n|=n^{-3}\to0$. The <squeeze theorem> gives $a_n\to0$, and every <subsequence> of a <convergent sequence> has the same <limit of a sequence>. Thus its <set> of <subsequential limits> is $\boxed{\{0\}}$.
For the second <sequence>, the four possible <sine> values give
$$
b_n=\begin{cases}
1,&n\equiv0,2\pmod4,\\
n,&n\equiv1\pmod4,\\
1/n,&n\equiv3\pmod4.
\end{cases}
$$
The even-indexed <subsequence> is constantly one, and $b_{4k+3}\to0$, so both zero and one occur as <subsequential limits>. A <convergent subsequence> is bounded, so it contains only finitely many terms with $n\equiv1\pmod4$. Of the remaining three residue classes, at least one occurs infinitely often. The corresponding further <subsequence> has <limit of a sequence> zero or one; the <subsequence> rule forces the original <convergent subsequence> to have that same <limit of a sequence>. Therefore the exact answer is $\boxed{\{0,1\}}$. Divergence to $+\infty$ along $n\equiv1\pmod4$ is not a finite <subsequential limit>.
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