5. (10 points) Let $[x]$ denote the greatest integer not exceeding $x$, and let $\{x\}=x-[x]$. Then the value of the expression $\left\{\frac{2012+1}{5}\right\}+\left\{\frac{2012+2}{5}\right\}+\left\{\frac{2012+3}{5}\right\}+\cdots+\left\{\frac{2012+2012}{5}\right\}$ is $\qquad$ .
【Analysis】According to the notation $[x]$ for the greatest integer not exceeding $x$, and $\{x\}=x-[x]$, each term of the original expression can be written as $\{x\} = x - [x]$, and then the calculation can be performed. Based on this, solve the problem.

【Solution】Solution: $\left\{\frac{2012+1}{5}\right\}+\left\{\frac{2012+2}{5}\right\}+\left\{\frac{2012+3}{5}\right\}+\cdots+\left\{\frac{2012+2012}{5}\right\}$,
$$
\begin{array}{l}
=\frac{2012+1}{5}-\left[\frac{2012+1}{5}\right]+\frac{2012+2}{5}-\left[\frac{2102+2}{5}\right]+\frac{2012+3}{5}-\left[\frac{2012+3}{5}\right]+\cdots+\frac{2012+2012}{5} \\
-\left[\frac{2102+2012}{5}\right], \\
=\frac{2012+1}{5}-402+\frac{2012+2}{5}-402+\frac{2012+3}{5}-403+\cdots+\frac{2012+2012}{5}-804, \\
=\frac{2010+3}{5}-402+\frac{2010+4}{5}-402+\frac{2015}{5}-403+\cdots+\frac{4020+4}{5}-804, \\
=402+\frac{3}{5}-402+\frac{4}{5}-402+403-403+\cdots+804+\frac{4}{5}-804, \\
=\frac{3}{5}+\frac{4}{5}+0+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}+0+\cdots \frac{4}{5}, \\
=\frac{3}{5}+\frac{4}{5}+(2012-2) \div 5 \times\left(0+\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right), \\
=\frac{3}{5}+\frac{4}{5}+2010 \div 5 \times 2, \\
=0.6+0.8+804, \\
=805.4 .
\end{array}
$$

Therefore, the answer is: 805.4.