2. Let $m$ be a real number, and consider the quadratic equation in $x$
$$
(m-2) x^{2}+(2 m-1) x+(m+2)=0
$$

The maximum value of the sum of the reciprocals of the two real roots is 8. Then the range of values for $m$ is $\qquad$
2. $m<-2$ or $-\frac{3}{2} \leqslant m \leqslant \frac{17}{4}$ and $m \neq 2$.

Let the two real roots of the equation be $x_{1}$ and $x_{2}$. By the problem, we have
$$
\left\{\begin{array}{l}
m-2 \neq 0, \\
\Delta=(2 m-1)^{2}-4(m-2)(m+2) \geqslant 0 .
\end{array}\right.
$$

Solving this, we get $m \leqslant \frac{17}{4}$ and $m \neq 2$.
By the relationship between roots and coefficients, we have
$x_{1}+x_{2}=-\frac{2 m-1}{m-2}, x_{1} x_{2}=\frac{m+2}{m-2}$.
Also, $\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}=-\frac{2 m-1}{m+2} \leqslant 8$, then
$$
\frac{8(m+2)+(2 m-1)}{m+2} \geqslant 0 \text {. }
$$

Thus, $m \geqslant-\frac{3}{2}$ or $m<-2$.
Therefore, the range of $m$ is $m<-2$ or $-\frac{3}{2} \leqslant m \leqslant \frac{17}{4}$ and $m \neq 2$.