1. Given that $x$ and $y$ are real numbers, and satisfy
$$
\left(x+\sqrt{x^{2}+2008}\right)\left(y+\sqrt{y^{2}+2008}\right)=2008 \text {. }
$$

Then the value of $x^{2}-3 x y-4 y^{2}-6 x-6 y+2008$
is $\qquad$
II, 1.2008.
From the given, we have
$$
x+\sqrt{x^{2}+2008}=\frac{2008}{y+\sqrt{y^{2}+2008}} .
$$

Rationalizing the denominator, we get
$$
x+\sqrt{x^{2}+2008}=\sqrt{y^{2}+2008}-y .
$$

Similarly, $y+\sqrt{y^{2}+2008}=\sqrt{x^{2}+2008}-x$.
$$
\begin{array}{l}
\text { (1) }+ \text { (2) gives } x+y=0 \text {. } \\
\text { Therefore, } x^{2}-3 x y-4 y^{2}-6 x-6 y+2008 \\
=(x+y)(x-4 y)-6(x+y)+2008 \\
=2008 \text {. }
\end{array}
$$