Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$ 
$$\left|\frac{3x+1}{x-2}\right|<1$$

$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{  and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{  and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{  and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is 
$${\frac{1}{4}}>{x} \text{  and } x>{\frac{-3}{2}}$$

When you multiply by $x-2$, you have to worry about whether it's negative, in which case the direction of the inequalities would be reversed.  So break things into two cases: $x>2$ and $x<2$.  
If $x>2$, you can do what you did or be a little more efficient:
$$-(x-2) < 3x+1 < x-2$$
$$-x +1 < 3x < x-3$$
$$1<4x \mbox{ and } 2x<-3$$
$$\frac{1}{4}<x \mbox{ and } x<-\frac{3}{2}$$
This is the empty set because no $x$ satisfies both inequalities.
Repeat for $x<2$ but reverse the inequalities when you multiply by it,
to get 
$$\frac{1}{4}>x \mbox{ and } x>-\frac{3}{2}$$
