17. (12 points) Given the function $f(x)$ for any real numbers $x, y$, it satisfies $f(x+y)=f(x)+f(y)-3$, and when $x>0$, $f(x)<3$.
(1) Is $f(x)$ a monotonic function on the set of real numbers $\mathbf{R}$? Explain your reasoning;
(2) If $f(6)=-9$, find $f\left(\left(\frac{1}{2}\right)^{2010}\right)$.
17. (1) For any $x_{1}, x_{2} \in \mathbf{R}$, and $x_{1} < x_{2}$, we have $x_{2} - x_{1} > 0$, so $f\left(x_{2}-x_{1}\right) < 3$, which means $f\left(x_{2}\right) < f\left(x_{1}\right)$.
Thus, $f(x)$ is a monotonically decreasing function on $\mathbf{R}$.
$$
\begin{array}{l}
\text { (2) From } f(6)=f(2)+f(4)-3 \\
=f(2)+(f(2)+f(2)-3)-3 \\
=3 f(2)-6=-9,
\end{array}
$$

we get $f(2)=-1$.
Also, from $f(2)=f(1)+f(1)-3$, we get $f(1)=1$.
For any $a \in \mathbf{R}$, it is clear that
$f(a)=2 f\left(\frac{a}{2}\right)-3 \Rightarrow f\left(\frac{a}{2}\right)=\frac{f(a)+3}{2}$.
Let $a_{1}=f(1)=1$,
$a_{n+1}=f\left(\left(\frac{1}{2}\right)^{n}\right)=\frac{f\left(\left(\frac{1}{2}\right)^{n-1}\right)+3}{2}=\frac{a_{n}+3}{2}$.
Let $b_{n}=a_{n+1}-a_{n}$. Then
$$
b_{n}=\frac{1}{2} b_{n-1}, b_{1}=a_{2}-a_{1}=1 .
$$

Thus, $b_{n}=\left(\frac{1}{2}\right)^{n-1}$, which means
$$
a_{n+1}-a_{n}=\left(\frac{1}{2}\right)^{n-1} \text {. }
$$

Therefore, $a_{n+1}=a_{1}+1+\frac{1}{2}+\cdots+\left(\frac{1}{2}\right)^{n-1}$
$$
=3-\left(\frac{1}{2}\right)^{n-1} \text {. }
$$

So, $f\left(\left(\frac{1}{2}\right)^{2010}\right)=a_{2011}=3-\left(\frac{1}{2}\right)^{2009}$.