$$\begin{array}{l} \text {-.1.D. } \\ N=\left\{x \left\lvert\, \frac{1}{2-x}>1\right., x \in \mathbf{R}\right\} \\ =\{x \mid 1<x<2, x \in \mathbf{R}\} . \end{array}$$

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, $a \geqslant \frac{1}{4}$.