33. Given that $h(x), k(x)$ are real-valued functions defined on $R$, and satisfy
$$
\begin{array}{l}
h(x) \cdot h(y)=\frac{k(x+y)-k(x-y)}{2}, \\
\quad k(x) \cdot k(y)=\frac{k(x+y) 2+k(x-y)}{2}. \\
x, y \in R .
\end{array}
$$

Find all possible values of $h(0)$ and $k(0)$.
Let $x=y=0$ in (1) and (2), we get
$$
h^{2}(0)=0 \text { and } k^{2}(0)=k(0) .
$$

Thus, $h(0)=0, k(0)=0$ or $k(0)=1$.
But $k(0)=0$ cannot hold. Otherwise, by setting $x=y$ in (1) and (2), we get
$$
\begin{array}{l}
h^{2}(x)=\frac{k(2 x)-k(0)}{2}=\frac{k(2 x)}{2}, \\
k^{2}(x)=\frac{k(2 x)+k(0)}{2}=\frac{k(2 x)}{2} .
\end{array}
$$

Thus, $h^{2}(x)=k^{2}(x)$.
$$
\therefore h(x)=k(x) \text { or } h(x)=-k(x) .(x \in \mathbb{R}),
$$

If $h(x)=k(x)$, then from (1) and (2) we can get
$$
\begin{array}{l}
\frac{k(x+y)-k(x-y)}{2} \cdot \frac{k(x+y)+k(x-y)}{2} . \\
\therefore k(x-y)=0 .
\end{array}
$$

That is, for any $x \in \mathbb{R}$, $k(x)=0$. This contradicts the assumption that $k(x)$ is a non-zero function.

If $h(x)=-k(x)$, then similarly we can get $k(x)=0$. This also contradicts the assumption that $k(x)$ is a non-zero function.

Therefore, we must have $h(0)=0, k(0)=1$. To verify that this satisfies the conditions, we can take $K(x)=\frac{1}{2}\left(x^{2}-e^{-x}\right), k(x)=\frac{1}{2}\left(e^{x}\right)$.