16. Given a function $f(x)$ defined on $(0,+\infty)$ that satisfies:
(1) For any $x \in(0, \infty)$, it always holds that $f(2 x)=2 f(x)$;
(2) When $x \in(1,2]$, $f(x)=2-x$.

The following conclusions are given: (1) For any $m \in \mathbf{Z}$, $f\left(2^{m}\right)=0$;
(2) The range of the function $f(x)$ is $[0,+\infty)$;
(3) There exists $n \in \mathbf{Z}$ such that $f\left(2^{n}+1\right)=9$;
(4) The necessary and sufficient condition for "the function $f(x)$ is monotonically decreasing on the interval $(a, b)$" is "there exists $k \in \mathbf{Z}$ such that $(a, b) \subseteq\left(2^{k}, 2^{k+1}\right)$". The correct conclusions are:
When $x \in\left(2^{k}, 2^{k+1}\right]$, $\frac{x}{2^{k}} \in(1,2] \Rightarrow f(x)=2^{k} f\left(\frac{x}{2^{k}}\right)=2^{k+1}-x \in\left[0,2^{k}\right)$, thus (1), (2), and (4) are correct. If (3) is correct, let $2^{n}+1 \in\left(2^{k}, 2^{k+1}\right] \Rightarrow f\left(2^{n}+1\right)=2^{k+1}-2^{n}-1=9 \Rightarrow 2^{k+1}-2^{n}=10$ $\Rightarrow k$ does not exist, leading to a contradiction. Therefore, the answer is (1)(2)(4).