calculate $f(500)=?$ and give an example of a function that meets the given conditions $f : \mathbb{R} \to \mathbb{R}$
$\forall_{x\in \mathbb{R}} f(x) * f(f(x))=1 (*) \\
f(1000)=999$
calculate $f(500)=?$
So $f(1000)*f(f(1000))=1 \to
999*f(999)=1 \to
f(999)= \frac{1}{999}$
Darboux. 
$\exists x_0 \in (999,1000)$
With (*) 
$ x=x_0$
$f(x_0)*f(f(x_0))=1 \to $
$500*f(500)=1 \to $
$f(500)=\frac{1}{500}$
I don't know how to give an example of a function that meets the given conditions.
 A: Any $y$ in the image of $f$ will have $y=f(x)$ for some $x \in \mathbb R$ and then satisfy $y \times f(y)=1$ so $f(y)=\frac1y$ and thus $f\left(\frac1y\right)=y$.  Let's call $f$'s  image $Y=\{y \in \mathbb R \mid \exists x \in \mathbb R: y=f(x)\}$ 
You now know that $0$ cannot be in $Y$.  Since $f(1000)=999$, you also know $1000$ and $\frac1{1000}$ cannot be in $Y$ and that $999$ and $\frac1{999}$ must be in $Y$.  Any such $f$ and $Y$ which meets these conditions with $f(y)=\frac1y$ for $y \in Y$ will be a solution. So $f(500)$ can potentially take any value apart from  $0$ or $1000$ or $\frac1{1000}$ or $500$ 
To construct an example, suppose you want $f(500)=k \not \in \left\{0 , 1000,\frac1{1000},500 \right\}$. Then consider 
$$   
f(x) = 
     \begin{cases}
       999 &\quad\text{if } x= 1000 \text{ or } x= \tfrac1{999}\\
       \tfrac1{999} &\quad\text{if } x= 999\\
       \tfrac1{k} &\quad\text{if } x= k\\
       k &\quad\text{otherwise.} \\ 
     \end{cases}$$ 
There will be many other possible examples
