Nine, (1) On the parabola $\mathrm{y}=\mathrm{x}^{2}-1$, find two distinct points such that these two points are symmetric with respect to the line $\mathbf{x}+\mathrm{y}=0$;
(2) For the parabola $y=2 x^{2}-1$ to always have two points symmetric with respect to $x+y=0$, find the range of $a$.
Nine, Solution
(1) Let the line perpendicular to $x+y=0$ be $y=x+b$. Solving the system of equations $\left\{\begin{array}{l}x+y=0 \\ y=x+b\end{array}\right.$, we get their intersection point as $\left(-\frac{b}{2}, \frac{b}{2}\right)$.
From $\left\{\begin{array}{l}y=x+b \\ y=x^{2}-1\end{array}\right.$, we get $x^{2}-x-(1+b)=0$.
If the coordinates of the two points are $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$, then according to the problem, we have: $\frac{x_{1}+x_{2}}{2}=-\frac{b}{2}$.
Since $x_{1}+x_{2}=1$, we have $b=-1$.
Solving the equation $x^{2}-x=0$, we get $x_{1}=0, x_{2}=1$, thus $y_{1}=-1, y_{2}=0$.
Therefore, the two symmetric points are: $(0,-1),(1,0)$.
(2) Substituting $y=x+b$ into $y=a x^{2}-1$, we get
$$
a x^{2}-x-(1+b)=0.
$$

From $\frac{1}{2 a}=-\frac{b}{2}$, we get $b=-\frac{1}{a}$, so we have
$$
a x^{2}-x-\left(1-\frac{1}{a}\right)=0.
$$

For the equation to always have two distinct real roots, it is necessary and sufficient that:
$$
\triangle=1+4 a\left(1-\frac{1}{a}\right)=4 a-3>0,
$$

hence $a>\frac{3}{4}$.