If x is different from 0, is |x| < 1 ?

OK. First of all: I marked "|x|<1" in red for (1), because you don't get that |x| < 1 from this statement. If it were so then then you'd have an YES answer right away (as the question asks exactly about the same thing "is |x|<1")

Next, "is \(|x|<1\)?" means "is \(-1<x<1\)?"

(1) x*|x|<x holds true in two cases for \(x<-1\) and \(0<x<1\):
-----(-1)----(0)----(1)----
So as you can see this one is not sufficient to answer the question.

(2) |x| > x. Absolute value property: when \(x\geq{0}\) then \(|x|=x\) and when \(x<{0}\) (so when \(x\) is negative) then \(|x|=-x\). For the second case \(|x|=positive>x=negative\). That' why \(|x|>x\) means that \(x<0\) (or as I wrote if \(x\) were positive we would have \(|x|=x\) not \(|x|>x\)). You can also consider a simple example: \(|-3|>-3\) --> \(-3<0\). Thus the range for (2) is:
-------------(0)--------
Also insufficient to answer the question.

(1)+(2) Intersection of the ranges from (1) and (2) is:
----(-1)----(0)----(1)---- --> \(x<-1\). Which makes the answer to the question "is \(-1<x<1\)?" NO.

Check Absolute Value chapter of Math Book for more on this topic: math-absolute-value-modulus-86462.html

Hope it's clear.