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.