(1) Let
$$\frac{x}{x-1} = a, \frac{y}{y-1} = b, \frac{z}{z-1} = c,$$
then
$$x = \frac{a}{a-1}, \quad y = \frac{b}{b-1}, \quad z = \frac{c}{c-1}.$$
Since $xyz = 1$, we have
$$abc = (a-1)(b-1)(c-1),$$
that is
$$a + b + c - 1 = ab + bc + ca.$$
Therefore
$$\begin{aligned} a^2 + b^2 + c^2 &= (a + b + c)^2 - 2(ab + bc + ca) \\ &= (a + b + c)^2 - 2(a + b + c - 1) \\ &= (a + b + c - 1)^2 + 1 \\ &\ge 1. \end{aligned}$$
So
$$\frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \ge 1.$$

(2) Take $(x, y, z) = \left(-\frac{k}{(k-1)^2}, k - k^2, \frac{k-1}{k^2}\right)$, $k$ is an integer, then $(x, y, z)$ is a triple of rational numbers, with $x$, $y$, $z$ each different from $1$. What is more, a different integer $k$ gives a different triple of rational numbers.
$$\begin{aligned} & \frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} \\ &= \frac{k^2}{(k^2 - k + 1)^2} + \frac{(k - k^2)^2}{(k^2 - k + 1)^2} + \frac{(k-1)^2}{(k^2 - k + 1)^2} \\ &= \frac{k^4 - 2k^3 + 3k^2 - 2k + 1}{(k^2 - k + 1)^2} = 1. \end{aligned}$$
So the problem is proved.

(1) By $xyz = 1$, let $p = x$, $q = 1$, $r = \frac{1}{y}$, then $x = \frac{p}{q}$, $y = \frac{q}{r}$, $z = \frac{1}{xy} = \frac{r}{p}$, where $p$, $q$, $r$ are different from each other. We have
$$\begin{aligned} \frac{x^2}{(x-1)^2} + \frac{y^2}{(y-1)^2} + \frac{z^2}{(z-1)^2} &\ge 1 \\ \Leftrightarrow \frac{p^2}{(p-q)^2} + \frac{q^2}{(q-r)^2} + \frac{r^2}{(r-p)^2} &\ge 1. \end{aligned} \quad ①$$
Let
$$a = \frac{p}{p-q}, \quad b = \frac{q}{q-r}, \quad c = \frac{r}{r-p},$$
after the substitution, ① reduces to $\sum a^2 \ge 1$. Since
$$\frac{-1 + a}{a} = \frac{q}{p}, \quad \frac{-1 + b}{b} = \frac{r}{q}, \quad \frac{-1 + c}{c} = \frac{p}{r},$$
therefore
$$\begin{aligned} & \frac{-1 + a}{a} \cdot \frac{-1 + b}{b} \cdot \frac{-1 + c}{c} = 1, \\ & 1 - \sum a + \sum ab = 0. \end{aligned} \quad ②$$
By ②, we get
$$1 - \sum a^2 = -(a + b + c - 1)^2 \le 0,$$
so $\sum a^2 \ge 1$. Hence ① holds.

(2) Let $b = \frac{t^2 + t}{t^2 + t + 1}$, $c = \frac{t + 1}{t^2 + t + 1}$, $a = -\frac{bc}{b + c}$, here $t$ can be any rational number except $0$ and $-1$. While $t$ varies, only finitely many of $t$ values can make $b$, $c$, $a$ be $1$. That is, there are infinitely many triples of rational numbers $a$, $b$, $c$ each different from $1$, satisfying $\sum a = \sum a^2 = 1$. By
$$(x, y, z) = \left( \frac{a}{a-1}, \frac{b}{b-1}, \frac{c}{c-1} \right),$$
(2) holds.