5. $a \leqslant-\frac{1}{2}$.

Let the points be $M\left(x_{0}, y_{0}\right), P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
\begin{array}{l}
l: x_{0} x-y_{0} y=1, \\
\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{x_{0}}{y_{0}},
\end{array}
$$

and $\left\{\begin{array}{l}x_{1}^{2}-y_{1}^{2}=0, \\ x_{2}^{2}-y_{2}^{2}=0 .\end{array}\right.$
$$
\text { Hence } \frac{y_{2}-y_{1}}{x_{2}-x_{1}} \cdot \frac{\frac{y_{1}+y_{2}}{2}}{\frac{x_{1}+x_{2}}{2}}=1 \text {. }
$$

Combining with equation (1), we get
$$
\frac{\frac{x_{1}+x_{2}}{2}}{\frac{y_{1}+y_{2}}{2}}=\frac{y_{0}}{x_{0}},
$$

which means $M$ is the midpoint of $P Q$.
Noting that,
$$
\begin{array}{l}
\overrightarrow{R P} \cdot \overrightarrow{R Q}=|\overrightarrow{R M}|^{2}-\frac{1}{4}|\overrightarrow{P Q}|^{2} \\
\geqslant \frac{1}{4} P O^{2}-\frac{1}{4} P Q^{2}=-\frac{1}{4} O Q^{2} .
\end{array}
$$

Since point $M$ is in the first quadrant, we know $O Q^{2}-\frac{1}{2}$.
Therefore, the range of $a$ is $a \leqslant-\frac{1}{2}$.
5. Given the hyperbola $\Gamma: x^{2}-y^{2}=1$ and a point $M$ in the first quadrant on $\Gamma$, a tangent line $l$ is drawn through $M$ to $\Gamma$, intersecting the asymptotes of $\Gamma$ at points $P$ and $Q$ ($P$ is in the first quadrant), and $R$ is on the same asymptote as $Q$. If $\overrightarrow{R P} \cdot \overrightarrow{R Q} \geqslant a$ always holds, then the range of values for $a$ is