$$
\begin{array}{l}
\text { Given } \mathrm{a}=\sqrt{3}+1, \mathrm{~b}=\sqrt{3}-1, \mathrm{~m} \angle \mathrm{C}=60^{\circ} \\
\therefore \quad \cos 60^{\circ}=\frac{(\sqrt{3}+1)^{2}+(\sqrt{3}-1)^{2}-\mathrm{c}^{2}}{2(\sqrt{3}+1)(\sqrt{3}-1)} \\
\qquad \frac{1}{2}=\frac{8-\mathrm{c}^{2}}{2(2)} \Rightarrow \mathrm{c}^{2}=6 \Rightarrow \mathrm{c}=\sqrt{6} \\
\text { By sine rule, } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \Rightarrow \frac{\sqrt{3}+1}{\sin \mathrm{A}}=\frac{\sqrt{6}}{\sin 60^{\circ}} \\
\therefore \quad \frac{\sqrt{3}+1}{\sin \mathrm{A}}=\frac{\sqrt{6}}{\left(\frac{\sqrt{3}}{2}\right)} \Rightarrow \sin \mathrm{A}=\frac{\sqrt{3}}{2} \times \frac{(\sqrt{3}+1)}{\sqrt{6}} \Rightarrow \sin \mathrm{A}=\frac{\sqrt{3}+1}{2 \sqrt{2}} \\
\therefore \mathrm{A}=75^{\circ} \text { or }\left(180^{\circ}-75^{\circ}\right)=105^{\circ} \\
\therefore \mathrm{B}=180^{\circ}-\left(75^{\circ}+60^{\circ}\right)=45^{\circ} \text { or } 180^{\circ}-\left(105^{\circ}+60^{\circ}\right)=15^{\circ} \\
\text { By } \sin \mathrm{e} \text { rule, } \frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}} \Rightarrow \frac{\sqrt{3}-1}{\sin \mathrm{B}}=\frac{\sqrt{6}}{\sin 60^{\circ}} \\
\therefore \sin \mathrm{B}=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}-1}{\sqrt{6}}=\frac{\sqrt{3}-1}{2 \sqrt{2}} \Rightarrow \mathrm{B}=15^{\circ} \text { or }\left(180-15^{\circ}\right)=165^{\circ} \\
\text { Since } \mathrm{B} \neq 165^{\circ} \text { as } \mathrm{C}=60^{\circ} \text { given, we take } \mathrm{B}=15^{\circ} \\
\therefore \angle \mathrm{B}=15^{\circ}, \angle \mathrm{A}=105^{\circ} \Rightarrow \mathrm{A}-\mathrm{B}=90^{\circ}
\end{array}
$$