\[ \frac{4}{3} \cot^2 30^\circ + 3 \sin^2 60^\circ - 2 \csc^2 60^\circ - \frac{3}{4} \tan^2 30^\circ \] \[ = \frac{4}{3} (\sqrt{3})^2 + 3 \left(\frac{\sqrt{3}}{2}\right)^2 - 2 \left(\frac{2}{\sqrt{3}}\right)^2 - \frac{3}{4} \left(\frac{1}{\sqrt{3}}\right)^2 \] \[ = \frac{4}{3} \times 3 + 3 \times \frac{3}{4} - 2 \times \frac{4}{3} - \frac{3}{4} \times \frac{1}{3} \] \[ = 4 + \frac{9}{4} - \frac{8}{3} - \frac{1}{4} \] \[ = \frac{48 + 27 - 32 - 3}{12} = \frac{40}{12} = \frac{10}{3} = 3\frac{1}{3} \] (Answer)