In $\triangle A B C, a=2, b=\sqrt{6}, c=\sqrt{3}+1$
We know that,
$$
\begin{aligned}
\cos C & =\frac{a^2+b^2-c^2}{2 a b}=\frac{(2)^2+(\sqrt{6})^2-(\sqrt{3}+1)^2}{2 \cdot 2 \cdot \sqrt{6}} \\
& =\frac{4+6-3-1-2 \sqrt{3}}{4 \sqrt{6}} \\
& =\frac{6-2 \sqrt{3}}{4 \sqrt{6}}=\frac{2 \sqrt{3}(\sqrt{3}-1)}{4 \sqrt{6}} \\
\cos C & =\frac{\sqrt{3}-1}{2 \sqrt{2}} \\
\therefore \sin C & =\sqrt{\frac{1-\left(\frac{\sqrt{3}}{2 \sqrt{2}}\right)^2}{[\because \sin \theta}=\sqrt{\left.1-\cos ^2 \theta\right]}} \\
& =\sqrt{\frac{8-3-1+2 \sqrt{3}}{8}}=\sqrt{\frac{4+2 \sqrt{3}}{8}} \\
& =\sqrt{\frac{2+\sqrt{3}}{4}}=\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}
\end{aligned}
$$
Now, $\begin{aligned} \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{6+3+1+2 \sqrt{3}-4}{2 \sqrt{6}(\sqrt{3}+1)} \\ & =\frac{6+2 \sqrt{3}}{2 \sqrt{6}(\sqrt{3}+1)}=\frac{3+\sqrt{3}}{\sqrt{6}(\sqrt{3}+1)}=\frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{6}(\sqrt{3}+1)}\end{aligned}$
$$
\begin{aligned}
\cos A & =\frac{1}{\sqrt{2}} \\
\therefore \quad \sin A & =\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^2}=\sqrt{\frac{1}{2}}
\end{aligned}
$$
$\begin{aligned} & \text { Now, } \sin ^2 C-\sin ^2 A \\ & =\left(\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}\right)^2-\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{\sqrt{3}}{4}-\frac{1}{2}=\frac{\sqrt{3}}{4}\end{aligned}$