The inequality to be solved in Lang's Basic Mathematics is $x^2>1$ from which I've worked out:
$$x^2>1$$ $$\sqrt {x^2}> \sqrt 1$$ $${x>1} \,\text{or}\ \, {x<-1}$$
Which in my mind would be written:
$$x \in \mathbb{R},x\neq \{1, -1\}$$
$$\text{*Or per Michael Rozenberg's suggestion*:}\ $$ $$\lvert x\rvert >1$$ But which the textbook wrote as:
$$-1>x>1$$
Is that notation tradition? I assumed from: $$a>b>c \Rightarrow a>c$$ $$then$$ $$-1>x>1$$ $$-1>1$$
Which is obviously false.
On
You have done the algebra correctly; the answer is $x < -1$ or $x > 1$. But it would not be written $x \in \mathbb{R},x\neq \{1, -1\}$. This notation excludes $1$ and $-1$ but not the interval between them. So I would leave it as two inequalities.
Your book's notation combines the two inequalities into one, but makes it worse. Since in English we read left to right, the expectation is for the inequalities to hold all the way through. Generally, that notation works when $x$ is between the two end values. At best, it's sloppy to write this answer this way.
I like to think of this geometrically. Subtract $1$ from both sides of the inequality and you have $x^2 - 1 > 0$. Graph this on any graphing utility (here is a screenshot from desmos.com):
Clearly you can see two distinct sections (they are not continuous) where the graph is greater than $0$. These correspond to $x<-1$ and $x>1$. Because they are discontinuous, I would think the best way to express the answer is in a pair of inequalities, as we have above.
I think it's better to make the following.
$x^2>1\Leftrightarrow|x|>1$, which gives $x>1$ or $x<-1$.
Because $\sqrt{x^2}=|x|$.
About your second question.
$-1>x>1$ says $x>1$ and $x<-1$, which is absurd.
More things.
"," says "and".
From here it's better to write $x>1$ or $x<-1$ because $x>1$ , $x<-1$ is absurd again.