(a) The curve \(y = \frac{x+1}{x-1}\) has a vertical asymptote at \(x = 1\) and a horizontal asymptote at \(y = 1\). The branches are in the first and third quadrants.

(b) The curve \(y = \frac{|x|+1}{|x|-1}\) is symmetrical about \(x = 0\). The critical points are found by solving \(\frac{|x|+1}{|x|-1} = -2\), leading to \(|x| = \frac{1}{3}\). Thus, the set of values of \(x\) for which \(\frac{|x|+1}{|x|-1} < -2\) is \(-1 < x < -\frac{1}{3}, \frac{1}{3} < x < 1\).