Minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$ in $\Delta ABC$ 
In $\triangle ABC$, $\sin B=-\cos C$. Find the minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$.

According to the law of sines, $\begin{aligned}\frac{a^2+b^2}{c^2}=\frac{\sin^2A+\sin^2B}{\sin^2C}\end{aligned}$.
Solution $1$
Let $\sin B=-\cos C=k$, then $\sin B=k>0$, so $\cos C=-k<0$, meaning that $\begin{aligned}C>\frac\pi2\end{aligned}$. Thus $\begin{aligned}B<\frac\pi2\end{aligned}$, so $\cos B=\sqrt{1-k^2}$.
Now $\sin A=\sin(B+C)=k(-k)+\sqrt{1-k^2}\times\sqrt{1-k^2}=1-2k^2$. So $$\frac{\sin^2A+\sin^2B}{\sin^2C}=4-4k^2+\frac{10}{1-k^2}-13\ge2\sqrt{40}-13=4\sqrt{10}-13.$$
Solution $2$
From $\sin B=-\cos C>0$ we have $$B=C-\frac{\pi}{2}, \sin B=\sin \left(C-\frac{\pi}{2}\right)=-\cos C ,
\sin A=\sin (B+C)=\sin \left(2 C-\frac{\pi}{2}\right)=-\cos 2 C$$. So \begin{aligned}\frac{\sin ^{2} A+\sin ^{2} B}{\sin ^{2} C}&=\frac{\cos ^{2} 2 C+\cos ^{2} C}{\sin ^{2} C} \\
&=\frac{\left(1-2 \sin ^{2} C\right)^{2}+\left(1-\sin ^{2} C\right)}{\sin ^{2} C} \\
&=\frac{2+4 \sin ^{4} C-5 \sin ^{2} C}{\sin ^{2} C}=\frac{2}{\sin ^{2} C}+4 \sin ^{2} C-5 \\
& \geqslant 2 \sqrt{\frac{2}{\sin ^{2} C} \cdot 4 \sin ^{2} C}-5=4 \sqrt{2}-5,
\end{aligned}
 A: Everything should be correct in solution $2$ because AM-GM can be used as $\sin^2 C$ is always non-negative.
For solution $1$ however, after $\sin A = 1 - 2k^2$ your steps don't follow: something must have been gone catastrophically wrong there. Using the values that we obtained for $\sin A, \sin B$ and that $\sin^2 C = 1 - \cos^2 C$:
$$\frac{\sin^2A+\sin^2B}{\sin^2C}= \frac{(1 - 2k^2)^2 + k^2}{1-k^2} = \frac{1 - 3k^2 + 4k^4}{1 - k^2}.$$
After doing some synthetic division on $\frac{4x^2 - 3x + 1}{x - 1}$ (or polynomial long division if you prefer):
$$
\begin{array}{c|cc}
  1 & 4 & -3 & 1\\
  \ & \ & 4 & 1 \\
\hline
 \ & 4 & 1 & 2 \\
\end{array}
$$
and this implies $(4x^2 - 3x + 1) = (4x + 1)(x - 1) + 2$ or that $(4k^4 - 3k^2 + 1)$ $ = -(4k^2 + 1)(1 - k^2) + 2$. Thus we have:
$$\frac{1 - 3k^2 + 4k^4}{1 - k^2} = -(4k^2 + 1) + \frac{2}{1 - k^2} = 4 - 4k^2 + \frac{2}{1 - k^2} - 5.$$
instead of what you had there.
Now as $-1 ≤ \sin B ≤ 1$ and also $k \ne -1, 1$ due to the denominator, we can apply AM-GM since $1 - k^2$ will be non-negative.
This gives the minimum as $2 \sqrt{8} - 5 = 4 \sqrt{2} - 5$ like in your other answer. Thus solution $2$ is correct, but we can use AM-GM as well on your solution attempt $1$ with a bit more effort.
Addendum: the two functions which are in terms of $k$ or $\sin^2 C$ are remarkably similar. However, applying the transformation $x^2 \to 1 - x^2$ in both directions shows that the two functions do indeed have the same minimum, which is not obvious before polynomial division.
