By period, I assume you mean minimal period, otherwise there are inifinitely many periods.
Write $i=x_iy_i$, where every prime factor of $x_i$ divides $b$, and $\gcd(y_i, b)=1$. Then $f(i)=f(y_i)=ord_{y_i}(b)$, where $ord_{y_i}(b)$ is the multiplicative order of $b \pmod{y_i}$.
This is because:
$x_i$ does not affect the minimal period $f(i)$.
If $y_i \mid (b^n-1)$ for some $n$, then let $b^n-1=ky_i$, then
$$\frac{1}{y_i}=\frac{k}{b^n-1}=\frac{k}{b^n(1-b^{-n})}=\frac{k}{b^{n}}+\frac{k}{b^{2n}}+ \ldots$$
So $\frac{1}{y_i}$ has $n$ as a period. Since the smallest such $n$ is given by $ord_{y_i}(b)$, $f(y_i) \leq ord_{y_i}(b)$.
On the other hand, since $f(i)$ is a period, let $l$ be the number representing the repeating string, so we can write
$$\frac{1}{y_i}=\frac{l}{b^{f(i)}}+\frac{l}{b^{2f(i)}}+ \ldots=\frac{l}{b^{f(i)}(1-b^{-f(i)})}$$
Thus $b^{f(i)}-1=ly_i$, so $f(y_i) \geq ord_{y_i}(b)$.
Now back to your results.
1 is mostly correct, as we have $f(i)=f(y_i)=ord_{y_i}(b) \leq \varphi(y_i) \leq y_i-1 \leq i-1$ for $y_i>1$, and if $y_i=1$, $f(i)=1 \leq i-1$ for $i>1$. The only exception is $i=1$, as $f(i)=1>1-1$.
2 is partially correct. The $b^n-1$ part is correct, but not the $b^{n-1}$ part.
3 is correct. $y_i=y_{j_1}y_{j_2}$ so $f(i)=ord_{y_i}(b)=lcm(ord_{y_{j_1}}(b),ord_{y_{j_2}}(b))=lcm(f(j_1),f(j_2))$.
It suffices to prove the following statement: any non-mixed repeating decimal can be expressed as a fraction with denominator not divisible by $2$ or $5$.
Proof. Let $x$ be a real number with a non-mixed repeating decimal, given by
$$
0.\overline{x_1 x_2 x_3 \ldots x_n}
$$
for some $n$ and digits $x_i$. Then let
$$
A = 10^{n-1}x_1 + 10^{n-2} x_2 + \cdots + x_n < 10^n
$$
so that
\begin{align*}
10^nx
= x_1 x_2 x_3 \ldots x_n.\overline{x_1 x_2 x_3 \ldots x_n}
&= A + x
\end{align*}
Hence
$$
x = \frac{A}{10^n - 1} \\
$$
and the denominator is not divisible by $2$ or $5$.
Best Answer
Repeating decimals are simply geometric series which add up to
$$a + ar + ar^2 + \cdots = \frac {a}{1-r} $$
where $a$ and $r$ are rational numbers, so the result is rational.
For example:
$$\begin{align} 0.23\,23\,23\ldots &= 0.23 +0.00\,23+0.00\,00\,23 +\cdots \\[2ex] &= 0.23 + 0.23\left(\frac1{100}\right) + 0.23\left(\frac1{100}\right)^2 + \cdots \\[2ex] &= \frac{0.23}{1-\frac1{100}}=\frac {23}{99} \end{align}$$