1. In a geometric sequence, $a_{3}=\frac{8}{2}, S_{3}=\frac{9}{2}$. Find $a_{1}$ and q.
Solving from the given:
$$
\left\{\begin{array}{l}
\frac{9}{2}=\frac{a_{1}\left(1-q^{8}\right)}{1-q} \\
\frac{3}{2}=a_{1} q^{2}
\end{array}\right.
$$
$\Rightarrow\left\{\begin{array}{l}q=-\frac{1}{2} \\ a_{1}=6,\end{array}\right.$ or $\left\{\begin{array}{l}q=1, \\ a_{1}=\frac{3}{2}\end{array}\right.$.
The flaw in this notation is that it overlooks the premise $q \neq 1$ in the formula $\mathrm{Sn}=\frac{\mathrm{a}_{1}\left(1-\mathrm{q}^{*}\right)}{1-\mathrm{q}}$. It should be revised to:
When $q \neq 1$,
$$
\begin{array}{l}
\text { then, }\left\{\begin{array}{l}
\frac{9}{2}=\frac{a_{1}\left(1-q^{3}\right)}{1-q}, \\
\frac{3}{2}=a_{1} q^{2}
\end{array}\right. \\
\Longrightarrow\left\{\begin{array}{l}
q=-\frac{1}{2}, \\
a_{1}=6 .
\end{array}\right.
\end{array}
$$

When $q=1$, it is a constant sequence: $a_{1}:=2==\frac{3}{2}$, and $Z_{3}:=8 \varepsilon_{1}=\frac{9}{2}$, which is consistent with the given conditions.