7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2003}+2002(y-2)=1,
\end{array}\right.
$$

then $x+y=$ $\qquad$
7. Construct the function $f(t)=t^{2003}+2002 t$, it is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$, hence $x-1=2-y, x+y=3$. Therefore, the answer is 3.
8. Since $x$ is an acute angle, then $\cos x \neq 0$. Dividing both sides of the given equation by $\cos x$ yields $2 \sin x \cdot \tan x+\tan x-2 \sin x=3$, which simplifies to $(2 \sin x +1)(\tan x-1)=2$. Considering the function $f(x)=(2 \sin x+1)(\tan x-1)$ is strictly monotonically increasing in $\left(0, \frac{\pi}{2}\right)$, and $f(x)=2=f\left(\frac{\pi}{3}\right)$, hence $x=\frac{\pi}{3}$. Therefore, the answer is $-\frac{\pi}{3}$.