The recurring decimal $0.\overline{a_1\ldots a_n}$ is equal to
$$\frac{a_1\cdots a_n}{10^n-1}.$$
E.g., $x=0.\overline{285} = 0.285285285\cdots$, then
$$x = \frac{285}{10^3-1} = \frac{285}{999}.$$
That is, you get the periodic portion divided by a number that consists of as many $9$s as the length of the periodic portion.
There are many ways of seeing this; one is using geometric series. Another is to use some manipulations: if
$$x = 0.\overline{a_1\ldots a_n}$$
then
$$10^nx = a_1\ldots a_n . \overline{a_1\cdots a_n}$$
so
$$(10^n-1)x = 10^n x - x = a_1\cdots a_n.$$
The first "model solution" is using this: since $x = 0.\overline{6}$, then $x = \frac{6}{9}$ (the period has length $1$, so you get a single $9$ in the denominator.
When the periodic decimal does not start right after the decimal point, you need to shift it a bit first. So for example, if you had
$$ x = 0.1\overline{285} = 0.1285285285\ldots,$$
then first we take $10 x = 1.\overline{285}$, then proceed as before:
$$\begin{align*}
10^3(10 x) &= 1285.\overline{285}\\
10x &= 1.\overline{285}\\
10x(10^3-1) &= 1284\\
x(9990)&= 1284\\
x &= \frac{1284}{9990}.
\end{align*}$$
The second model solution uses this method.
Added. For the series method, in case anyone is interested, suppose that $x$ is of the form $x=0.\overline{a_1\cdots a_n}$. This means, explicitly, that
$$ x = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{(10^n)^k} = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{10^{nk}} = a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}}.$$
This is a geometric series, with initial term $\frac{1}{10^{n}}$ and common ratio $\frac{1}{10^n}$, so it converges. A geometric series with initial term $a$ and common ratio $r$, $|r|\lt 1$, converges to
$$\frac{a}{1 - r},$$
so we have
$$\begin{align*}
x &= a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}} \\
&= a_1\cdots a_n\left(\frac{\frac{1}{10^n}}{1 - \frac{1}{10^n}} \right)\\
&= a_1\cdots a_n\left(\frac{\quad\frac{1}{10^n}\quad}{\quad\frac{10^n-1}{10^n}\quad}\right)\\
&= a_1\cdots a_n\left(\frac{1}{10^n-1}\right) = \frac{a_1\cdots a_n}{10^n-1}\\
&= \frac{a_1\cdots a_n}{\underbrace{9\cdots 9}_{n\text{ digits}}}.
\end{align*}$$
And similarly if you have to "shift" the decimal before you get to the period; you simply add enough $0$s to the $9$s in the denominator to account for the shift.