Question 32, Given a strictly monotonic function $f(x)$ defined on $\mathrm{R}$ that satisfies $f(x+y)=f(x)+f(y)$, and $f(1)=2$,
(1) Prove that $f(x)$ is an odd function;
(2) When $t>2$, the inequality $f\left(k \cdot \log _{2} t\right)+f\left(\log _{2} t-\left(\log _{2} t\right)^{2}-2\right)<0$ always holds, find the range of real number $k$.

Question 32, Solution: (1) Let $x=y=0$, then $f(0)=f(0)+f(0)$, hence $f(0)=0$. Now let $y=-x$, then $f(0)=f(x)+f(-x)$, hence $f(-x)=-f(x)$, so $f(x)$ is an odd function.
(2) Since $f(0)=0, f(1)=2$, and $f(x)$ is monotonic on $\mathbb{R}$, $f(x)$ is strictly increasing on $\mathbb{R}$. Noting that $f(x)$ is an odd function and $t>2$, we have
$$
\begin{array}{l}
f\left(k \cdot \log _{2} t\right)+f\left(\log _{2} t-\left(\log _{2} t\right)^{2}-2\right)<0 \\
\Leftrightarrow f\left(k \cdot \log _{2} t\right)<-f\left(\log _{2} t-\left(\log _{2} t\right)^{2}-2\right) \\
\Leftrightarrow f\left(k \cdot \log _{2} t\right)<f\left(\left(\log _{2} t\right)^{2}-\log _{2} t+2\right) \\
\Leftrightarrow k \cdot \log _{2} t<\left(\log _{2} t\right)^{2}-\log _{2} t+2 \\
\Leftrightarrow k<\log _{2} t+\frac{2}{\log _{2} t}-1
\end{array}
$$

Noting that $\log _{2} t \in[1,+\infty)$, the range of $\log _{2} t+\frac{2}{\log _{2} t}-1$ is $[2 \sqrt{2}-1,+\infty)$. Therefore, the range of $k$ that satisfies the condition is $(-\infty, 2 \sqrt{2}-1)$.