Answer
$f[g(x)]=\displaystyle \frac{8\sqrt{3-x}}{3-x}$
$g[f(x)]=\displaystyle \frac{\sqrt{3x^{2}-8x}}{x}$
Work Step by Step
In the expression for f(x), replace x with g(x)
$f[g(x)]=\displaystyle \frac{8}{g(x)}$
$=\displaystyle \frac{8}{\sqrt{3-x}}$
... rationalize the denominator
$=\displaystyle \frac{8}{\sqrt{3-x}}\cdot\frac{\sqrt{3-x}}{\sqrt{3-x}}$
$=\displaystyle \frac{8\sqrt{3-x}}{3-x}$
In the expression for g(x), replace x with f(x)
$g[f(x)]=\sqrt{3-f(x)}$
$=\sqrt{3-\dfrac{8}{x}}$
$=\sqrt{\dfrac{3x-8}{x}}$
... rationalize the denominator
$=\displaystyle \frac{\sqrt{3x-8}}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}}$
$=\displaystyle \frac{\sqrt{3x^{2}-8x}}{x}$