a. (4-2$\sqrt{3}$ ).(2+$\sqrt{3}$ )
= 4.2 + 4$\sqrt{3}$ - 2$\sqrt{3}$.2 - 2$\sqrt{3}$.$\sqrt{3}$
= 8 + 4$\sqrt{3}$ - 4$\sqrt{3}$ - 6
= 2
b. $\frac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}$ - $\frac{3-\sqrt{3}}{\sqrt{3}-1}$
= ($\sqrt{6}$ -$\sqrt{3}$).($\sqrt{2}$ +1) - $\frac{(3-\sqrt{3}).(\sqrt{3}+1)}{2}$
= $\sqrt{12}$ + $\sqrt{6}$ - $\sqrt{6}$ - $\sqrt{3}$ - $\frac{3\sqrt{3}+3-3-\sqrt{3}}{2}$
= 2$\sqrt{3}$ - $\sqrt{3}$- $\frac{2\sqrt{3}}{2}$
= 2$\sqrt{3}$ - $\sqrt{3}$ - $\sqrt{3}$
= 0
