2. Given the function $f(x)=\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right) \cos x$ $+b \sin x+6$ (where $a$ and $b$ are constants, and $a>1$), $f\left(\lg \log _{8} 1000\right)=8$. Then the value of $f(\lg \lg 2)$ is ( ).
(A) 8
(B) 4
(C) -4
(D) depends on the values of $a$ and $b$
2. (B).

Let $F(x)=f(x)-6$.
It is not hard to prove that $F(-x)=-F(x)$, i.e., $F(x)$ is an odd function.
Then
$$
\begin{array}{l}
F\left(-\lg \log _{8} 1000\right)=-F\left(\operatorname{lglog}_{8} 1000\right), \\
F(\operatorname{lglg} 2)=-F(\operatorname{lglog} 8000), \\
f(\lg 22)-6=-\left[f\left(\operatorname{lglog}_{8} 1000\right)-6\right]
\end{array}
$$

Therefore, $f(\operatorname{lglg} 2)=-f(\lg \log 81000)+12=4$.