(a) $f(x)=-\frac{4}{x}, x \neq 0$ \[ \begin{array}{l} g(x)=-\frac{4}{x}, x \neq 0 \\ f(g(x))=\square \\ g(f(x))=\square \end{array} \] $f$ and $g$ are inverses of each other $f$ and $g$ are not inverses of each other

Step 1 ：First, we calculate the value of $g(x)$, which is $g(x) = -\frac{4}{x}$, so $g(2) = -\frac{4}{2} = -2$

Step 2 ：Then, we substitute $g(2)$ into $f(x)$, so $f(g(2)) = f(-2) = -\frac{4}{-2} = 2$

Step 3 ：Next, we calculate the value of $f(x)$, which is $f(x) = -\frac{4}{x}$, so $f(2) = -\frac{4}{2} = -2$

Step 4 ：Then, we substitute $f(2)$ into $g(x)$, so $g(f(2)) = g(-2) = -\frac{4}{-2} = 2$

Step 5 ：Finally, we can see that $f(g(2)) = g(f(2)) = 2$, so $f$ and $g$ are inverses of each other