$$
\begin{array}{l}
\text { 8. Given } x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R} \text {, and } x^{3}+\sin x-2 a=0,4 y^{3}+ \\
\sin y \cos y+a=0, \cos (x+2 y)=
\end{array}
$$
8. 1 .

Let $f(t)=t^{3}+\sin t$, then $f(t)$ is an odd function and increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, and $f(x)=x^{3}+$ $\sin x=2 a . f(2 y)=8 y^{3}+2 \sin y \cos y=2\left[4 y^{3}+\sin y \cos y\right]=-2 a, f(x)=-f(2 y)=$ $f(-2 y) . x y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right],-2 y \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, according to monotonicity $x=-2 y$, hence $x+2 y=0$, $\cos (x+2 y)=1$