5. If the discrete random variables $X, Y$ satisfy
$$
2 \leqslant X \leqslant 3 \text {, and } X Y=1 \text {, }
$$

then the maximum value of $\mathrm{E}(X) \mathrm{E}(Y)$ is
5. $\frac{25}{24}$.

Let the possible values of the random variable $X$ be $x_{1}, x_{2}, \cdots, x_{n}$, with corresponding probabilities $p_{1}, p_{2}, \cdots, p_{n}$. Set
$$
\begin{array}{l}
m_{1}=\min \left\{x_{i} p_{i}\right\}, M_{1}=\max \left\{x_{i} p_{i}\right\}, \\
m_{2}=\min \left\{\frac{p_{i}}{x_{i}}\right\}, M_{2}=\max \left\{\frac{p_{i}}{x_{i}}\right\},
\end{array}
$$

where $i=1,2, \cdots, n$.
$$
\begin{array}{l}
\text { Hence } \mathrm{E}(X) \mathrm{E}(Y)=\left(\sum_{i=1}^{n} x_{i} p_{i}\right)\left(\sum_{i=1}^{n} \frac{p_{i}}{x_{i}}\right) \\
\leqslant \frac{1}{4}\left(\sum_{i=1}^{n} p_{i}\right)^{2}\left(\sqrt[4]{\frac{M_{1} M_{2}}{m_{1} m_{2}}}+\sqrt[4]{\frac{m_{1} m_{2}}{M_{1} M_{2}}}\right)^{2} \\
\leqslant \frac{1}{4} \times \frac{25}{6}=\frac{25}{24} .
\end{array}
$$