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We first substitute for $c$ to get $$\begin{array}{rl}\frac{a+2}{a+1}\cdot \frac{b-1}{b-2}\cdot \frac{c+8}{c+6}& =\frac{a+2}{a+1}\cdot \frac{b-1}{b-2}\cdot \frac{(b-10)+8}{(b-10)+6}\\ & =\frac{a+2}{a+1}\cdot \frac{b-1}{b-2}\cdot \frac{b-2}{b-4}.\end{array}$$ Since the denominators are not zero, we can cancel the $(b-2)$s to get $$\frac{a+2}{a+1}\cdot \frac{b-1}{b-4}.$$ Now, by the substitution $b=a+2$, this becomes $$\frac{a+2}{a+1}\cdot \frac{(a+2)-1}{(a+2)-4}=\frac{a+2}{a+1}\cdot \frac{a+1}{a-2}.$$ We can cancel as before to get $$\frac{a+2}{a-2},$$ which is equal to $\frac{4+2}{4-2}=\frac{6}{2}=\boxed{3}$, since $a=4$.

We could also solve for $b$ and $c$ before simplifying. Since $a=4$, we have $$b=a+2=4+2=6,$$ and then $$c=b-10=6-10=-4.$$ The expression then becomes $$\begin{array}{rl}\frac{a+2}{a+1}\cdot \frac{b-1}{b-2}\cdot \frac{c+8}{c+6}& =\frac{4+2}{4+1}\cdot \frac{6-1}{6-2}\cdot \frac{-4+8}{-4+6}\\ & =\frac{6}{5}\cdot \frac{5}{4}\cdot \frac{4}{2}\\ & =\frac{6}{2}=\boxed{3}.\end{array}$$