8. The domain of the function $f(x)$ is $\mathbf{R}$, and for any $x_{1}, x_{2} \in \mathbf{R}$, it holds that $f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)$. When $x \geqslant 0$, the function $f(x)$ is increasing.
(1) Prove that the function $y=-(f(x))^{2}$ is increasing on $(-\infty, 0]$ and decreasing on $[0,+\infty)$.
(2) If $f(1)=2$, and the real number $a$ satisfies the inequality $f\left(2 a^{2}-1\right)+2 f(a)-6<0$, find the range of values for $a$.
8. (1) Let $x_{1}=x_{2}=0$, then $f(0)=0$; let $x_{1}=x, x_{2}=-x$, we get $f(-x)=-f(x)$, hence $f(x)$ is an odd function. And when $x \geqslant 0$, $f(x)$ is an increasing function, so when $x \leqslant 0$, $f(x)$ is also an increasing function. Therefore, $f(x)$ is increasing on $\mathbf{R}$.

Suppose $x_{1}, x_{2} \in[0,+\infty)$, and $x_{1}f\left(x_{1}\right)$, thus $f\left(x_{1}\right)+f\left(x_{2}\right)>0, y_{1}>y_{2}$. $y=-(f(x))^{2}$ is decreasing on $[0,+\infty)$, similarly, it can be proven that $y=-(f(x))^{2}$ is increasing on $(-\infty, 0]$.
(2) Given $f(1)=2$, we have $f(2)=4, f(3)=6$. Therefore, $f\left(2 a^{2}-1\right)+2 f(a)-6<0 \Leftrightarrow f\left(2 a^{2}-1\right)+f(2 a)<f(3) \Leftrightarrow f\left(2 a^{2}+2 a-1\right)<f(3) \Leftrightarrow 2 a^{2}+2 a-1<3 \Leftrightarrow a \in(-2,1)$.