Given angle $\mathrm{A}=\frac{\pi}{3}$ and $\mathrm{B}=\frac{\pi}{4}$ In $\triangle \mathrm{ABC}$,
$$
\begin{aligned}
& \angle \mathrm{C}=\pi-(\mathrm{A}+\mathrm{B}) \\
& =\pi-\left(\frac{\pi}{3}+\frac{\pi}{4}\right) \\
& =\pi-\frac{7 \pi}{12}=\frac{5 \pi}{12}
\end{aligned}
$$
Apply sine law,
$$
\begin{aligned}
& \frac{\sin \mathrm{A}}{\mathrm{a}}=\frac{\sin \mathrm{B}}{\mathrm{b}}=\frac{\sin \mathrm{C}}{\mathrm{c}} \\
& \frac{\sin \left(\frac{\pi}{3}\right)}{\mathrm{a}}=\frac{\sin \left(\frac{\pi}{4}\right)}{\mathrm{b}}=\frac{\sin \left(\frac{5 \pi}{12}\right)}{\mathrm{c}} \\
& \frac{\sqrt{3}}{2 \mathrm{a}}=\frac{1}{\sqrt{2} \mathrm{~b}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}} \\
& \text { Compare } \frac{\sqrt{3}}{2 \mathrm{a}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}}
\end{aligned}
$$


Similarly, $\frac{1}{\sqrt{2} \mathrm{~b}}=\frac{\sqrt{3}+1}{2 \sqrt{2} \mathrm{c}}$

$$
\begin{aligned}
& \text { Take, } \frac{a^2}{c^2}-\frac{b^2}{c^2}=\frac{6}{(\sqrt{3}+1)^2}-\frac{4}{(\sqrt{3}+1)^2} \\
& \frac{2}{(3+1+2 \sqrt{3})}=\frac{2}{4+2 \sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} \\
& \frac{a^2-b^2}{c^2}=2-\sqrt{3}
\end{aligned}
$$
So, correct option is (a).