Let $f(x)=\frac{1}{x-2}$ and $g(x)=\frac{2}{x}+2$. Then $(f \circ g)(x)=$ \[ (g \circ f)(x)= \]

Step 1 ：Let \(f(x)=\frac{1}{x-2}\) and \(g(x)=\frac{2}{x}+2\).

Step 2 ：The composition of two functions, \(f\) and \(g\), denoted as \((f \circ g)(x)\) or \(f(g(x))\), is a function that applies \(g\) to its input, and then \(f\) to the result.

Step 3 ：Similarly, \((g \circ f)(x)\) or \(g(f(x))\) is a function that applies \(f\) to its input, and then \(g\) to the result.

Step 4 ：To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\), and to find \((g \circ f)(x)\), we need to substitute \(f(x)\) into \(g(x)\).

Step 5 ：The composition of the functions \(f\) and \(g\) is given by \((f \circ g)(x) = \frac{x}{2}\) and \((g \circ f)(x) = 2x - 2\).

Step 6 ：Therefore, we have \((f \circ g)(x) = \boxed{\frac{x}{2}}\) and \((g \circ f)(x) = \boxed{2x - 2}\).