1. If the set $M=\left\{x \left\lvert\, x^{2}-\frac{5}{4} x+a1\right., x \in \mathbf{R}\right\}
$

is an empty set, then the range of values for $a$ is ( ).
(A) $a \leqslant-\frac{3}{2}$
(B) $a \geqslant-\frac{3}{2}$
(C) $a \leqslant \frac{1}{4}$
(D) $a \geqslant \frac{1}{4}$
$$
\begin{array}{l}
\text {-.1.D. } \\
N=\left\{x \left\lvert\, \frac{1}{2-x}>1\right., x \in \mathbf{R}\right\} \\
=\{x \mid 1<x<2, x \in \mathbf{R}\} .
\end{array}
$$

From $M \cap N=\varnothing$, we get that $x^{2}-\frac{5}{4} x+a<0$ has no solution in $1<x<2$, which means $x^{2}-\frac{5}{4} x+a \geqslant 0$ always holds in $1<x<2$. Therefore, it must be true that
$$
\begin{array}{l}
a \geqslant-\left(x^{2}-\frac{5}{4} x\right) \\
=-\left(x-\frac{5}{8}\right)^{2}+\frac{25}{64}(1<x<2) .
\end{array}
$$

Thus, the range of $f(x)=-\left(x-\frac{5}{8}\right)^{2}+\frac{25}{64}(1<x<2)$ is $\left(-\frac{3}{2}, \frac{1}{4}\right)$. Therefore, $a \geqslant \frac{1}{4}$.