Let $f(x)=2-|x-3|, 1 \le x \le 5$ and for rest of the values $f(x)$ can be obtained by using the relation $f(5x)=\alpha\, f(x)\forall\, x \in R$.
The value of $f(2007)$ taking $\alpha = 5$, is: 

$f\left( x \right) =2-\left| x-3 \right| ,1\le x\le 5$

$f\left( 1 \right) =0$

$f\left( 2 \right) =1$

$f\left( 3 \right) =2$

$f\left( 4 \right) =1$

$f\left( 5 \right) =0$

$f\left( 2007 \right) =\alpha f\left( \cfrac { 2007 }{ 5 } \right)$

$={ \alpha }^{ 2 }f\left( \cfrac { 2007 }{ 25 } \right)$

$={ \alpha }^{ 3 }f\left( \cfrac { 2007 }{ 125 } \right)$

$={ \alpha }^{ 4 }f\left( \cfrac { 2007 }{ 625 } \right)$

$={ \alpha }^{ 5 }f\left(\cfrac { 2007 }{ 3125 } \right)$

$\cfrac { 2007 }{ 3125 } <1\therefore$ Rejected

$\quad f\left( 2007 \right) ={ \alpha }^{ 4 }f\left(\cfrac { 2007 }{ 625 } \right)$

$\cfrac { 2007 }{ 625 } \approx 3$

$f(2007)={ \alpha }^{ 4 }f(3)$

$={ 5 }^{ 4 }\times 2$

$=1250$