2. Given that the odd function $y=f(x)$ is decreasing on its domain $[-1,1]$, and $f(1-a)+f(1-a)^{2}>0$, find the range of the real number $a$.
2. Since the domain of $f(x)$ is $[-1,1]$, we have $0 \leqslant a \leqslant \sqrt{2}$.

Given that $f(x)$ is an odd function, it follows that $f\left(1-a^{2}\right)=-f\left(a^{2}-1\right)$.
From $f(1-a)+f\left(1-a^{2}\right)>0$, we get $f(1-a)>f\left(a^{2}-1\right)$.
Since $f(x)$ is a decreasing function, it follows that $1-a1$. Therefore, $a \in(1, \sqrt{2}]$.