17. Let the function $f(x)$ be monotonic on $(-\infty,+\infty)$, and satisfy $f(x+y)=f(x)+f(y)$ and $f(1)=$ -1. If for any positive number $t$ there is $f\left(k \log _{2} t\right)+f\left(\log _{2} t-\log _{2}^{2} t-2\right)>0$ always holds, find the range of $k$.

17. Let $x=1, y=0$, then $f(1)=f(1)+f(0)$,

so $f(0)=0$. Also, $f(1)=-1$, so $f(0)>f(1)$.
Since $f(x)$ is monotonic on $(-\infty,+\infty)$,
$f(x)$ is a decreasing function on $\mathbf{R}$, and $f(0)=0$,
hence $f(x)>0$ for all $x \in (-\infty, 0)$.
Therefore, $f\left(k \log _{2} t\right)+f\left(\log _{2} t-\log _{2}^{2} t-2\right)=f\left[-\log _{2}^{2} t+(k+1) \cdot \log _{2} t-2\right]>0$ always holds
$\Leftrightarrow-\log _{2}^{2} t+(k+1) \cdot \log _{2} t-2<0$ always holds
$\Leftrightarrow \Delta=(k+1)^{2}-8<0 \Leftrightarrow-1-2 \sqrt{2}<k<-1+2 \sqrt{2}$.
Thus, the range of $k$ is $(-1-2 \sqrt{2},-1+2 \sqrt{2})$.