Is it true that $\left(-\frac{1}{64}\right)^{-\frac 43}=256$? [duplicate]

You have put your finger precisely on the statement that is incorrect.

There are two competing conventions with regard to rational exponents.

The first convention is to define the symbol $a^x$ for $a > 0$ only. The symbol $\sqrt[n]{a}$ is defined for negative values of $a$ so long as $n$ is odd, but according to this convention, one wouldn't write $a^{1/n}$, for instance.

In defining $a^{p/q}$ to be $(\sqrt[q]{a})^p$, the author you quoted chose the fraction $p/q$ to be in lowest form so that the definition would be unambiguous. For example, $a^{10/15}$ is defined to be $(\sqrt[3]{a})^2$. However, it is preferable to define $a^{p/q}$ to be $(\sqrt[q]{a})^p$ in all cases and to prove that this definition is independent of the particular representation chosen for $p/q$; this is what more rigorous books tend to do. That is, you prove that if $p/q = r/s$, then $(\sqrt[q]{a})^p = (\sqrt[s]{a})^r$. There is no mention of lowest form.

The competing convention is to also allow $a^x$ to be defined for all $a \ne 0$ and all rational numbers $x = p/q$ that have at least one representation with an odd denominator. You then prove that $(\sqrt[q]{a})^p$ is independent of the particular representation $p/q$ chosen, so long as the denominator is odd. Thus you can write $a^{3/5} = (\sqrt[5]{a})^3 = (\sqrt[15]{a})^{9} = a^{9/15}$. All of that is fine. However, you cannot write $a^{6/10} = (\sqrt[10]{a})^6$, or even $a^{6/10} = \sqrt[10]{a^6}$. The number $a^{6/10}$ is well-defined, but to write down its definition, you must first select a fraction equivalent to $6/10$ that has an odd denominator, which could be $3/5$ or $9/15$ or something else. For $a^{1/2}$, this can't be done at all, so $a^{1/2}$ is undefined for $a < 0$.

The rules for exponents break down if you start allowing $a < 0$ and exponents that can't be written with an odd denominator. For example, the rule $a^{xy} = (a^x)^y$ is valid, but only so long as $x$ and $y$ are both rational numbers that can be written with an odd denominator. This is not the case if you write $a^1 = (a^2)^{1/2}$, despite the fact that both sides of the equation are defined since $a^2 > 0$.

Edit Reading the paper by Tirosh and Even, I was surprised to learn this matter has drawn serious attention from math educators.

A long time ago, I assumed that, apart from complex extensions, $a^x$ for non-integer $x$ should be defined only for $a > 0$. I reasoned that it made no sense to have a function $(-2)^x$ defined only for rational numbers $x$ with odd denominator. I objected strenuously to notations like $(-8)^{1/3}$.

But that was before I taught a calculus class, which is when I realized why some textbook authors are so happy to define $a^x$ for $a < 0$, following the second convention. The reason is that the formula $\frac{d}{dx}(x^r) = rx^{r-1}$ is perfectly valid for $x < 0$ and $r$ with odd denominator.

$-1 = (-1)^1 = (-1)^\frac{2}{2} = (-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2} = (1)^\frac{1}{2} = \sqrt{1} = 1$

The thing that looks the most suspect in my example above is the 4th equality, $(-1)^{2 \cdot \frac{1}{2}} = ((-1)^2)^\frac{1}{2}$, which seems to violate the spirit of Ratti's definition of rational exponents ("no common factors")... but technically, that translation from rational exponent to radical expression was not used as this point.

The 4th equality is indeed suspect, but not for the reason you suggest. It is an application the 2nd property of rational exponents that you list above:

If $r$ and $s$ are rational numbers and $a$ is a real number, then we have: $$(a^r)^s = a^{r\cdot s}$$

provided that all expressions used are defined.

More formally and less ambiguous would be:

$$\forall r,s \in \mathbb{Q}\colon \forall a \in \mathbb{R}\colon [a^r\in \mathbb{R} \land a^s\in \mathbb{R} \implies (a^r)^s=a^{r\cdot s}]$$

This statement makes it clear that we cannot infer $((-1)^2)^\frac{1}{2}=(-1)^{2 \times \frac{1}{2}}$ as in the "paradox" because $(-1)^\frac{1}{2} \notin \mathbb R$, i.e. because $(-1)^\frac{1}{2}$ is not defined.

That both restrictions are necessary can be seen from the fact that we must have $a^{r\cdot s}=a^{s\cdot r}=(a^s)^r=(a^r)^s$. If we had $a^s \notin \mathbb{R}$, we could not make this substitution.

With this in mind, we could restate the rule as follows:

$$\forall r,s \in \mathbb{Q}\colon \forall a \in \mathbb{R}\colon [a^r\in \mathbb{R} \land a^s\in \mathbb{R} \implies a^{r\cdot s}=(a^r)^s=(a^s)^r]$$

Though it has nothing to do with resolving the paradox, we may also need to define $x^\frac{1}{n}$ as follows:

$\forall x,y\in \mathbb{R}\colon\forall n\in \mathbb{N}\colon [Odd(n)\lor Even(n) \land n\neq 0 \land y\geq 0\implies [x^\frac{1}{n} =y\iff x=y^n ]]$

Using this rule, we could infer that $4^\frac{1}{2}=2$, but not $4^\frac{1}{2}=-2$.

BTW, as far as $\frac{m}{n}$ having to be in lowest terms, the definition given seems a bit sloppy. It cannot be, for example, that $4^\frac{2}{4}$ is undefined when $4^\frac{2}{4}= 4^\frac{1}{2}$ by substitution of $\frac{2}{4}=\frac{1}{2}$. I really don't think this notion can be source of the paradox.