解：?$(1)$?∵一元二次方程?$2x2-3x-1=0$?
的兩根分別為?$m$?、?$n$?
∴?$m+n=\frac {3}{2}$?，?$mn=-\frac {1}{2}$?
∴?$\frac {n}{m}+\frac m{n}=\frac {n2+\mathrm {m^2}}{mn}=\frac {(m+n)2-2mn}{mn}$?
?$=\frac {(\frac {3}{2})2-2×(-\frac {1}{2}) }{-\frac 12}=-\frac {13}{2}$?
?$(2)$?∵實(shí)數(shù)?$s$?、?$t $?滿足?$2s2-3s-1=0$?，
?$2t2-3t-1=0$?，?$s≠t$?
∴?$s $?與?$t $?可看作方程?$2x2-3x-1=0$?的兩個(gè)實(shí)數(shù)根
∴?$s+t=\frac {3}{2}$?，?$st=-\frac {1}{2}$?
∴?$(s-t)2=(s+t)2-4st=(\frac {3}{2})2-4×(-\frac {1}{2})=\frac {17}{4}$?
∴?$s-t=±\frac {\sqrt {17}}2$?
∴?$\frac {1}{s}-\frac {1}{t}=\frac {t-s}{st}=\frac {-(s-t)}{st}=± \sqrt {17}$?
