8. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+\cdots+2012}\right)$ $=$

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8. $\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+\cdots+2012}\right)$ $=$
8. $\frac{1007}{3018}$.

Given $1-\frac{1}{1+2+\cdots+n}=\frac{(n-1)(n+2)}{n(n+1)}$, then
$$
\begin{array}{l}
\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right) \cdots\left(1-\frac{1}{1+2+\cdots+2012}\right) \\
=\frac{1 \times 4}{2 \times 3} \cdot \frac{2 \times 5}{3 \times 4} \cdots \cdots \frac{2011 \times 2014}{2012 \times 2013} \\
=\frac{2014}{3 \times 2012}=\frac{1007}{3018} .
\end{array}
$$