解：?$(1)$?由題意，對(duì)于方程?$x^2-6x + 2m - 1 = 0，$?
因?yàn)?$x_{1} = 1，$?根據(jù)根與系數(shù)的關(guān)系?$x_{1}+x_{2}=6，$??$x_{1}·x_{2}=2m - 1。$?
?$ $?則?$1 + x_{2} = 6，$?解得?$x_{2} = 5；$?
?$ 1×x_{2}=2m - 1，$?即?$5 = 2m - 1，$?解得?$m = 3。$?
?$ (2) $?解：存在。
根據(jù)題意，方程?$x^2-6x + 2m - 1 = 0$?中，?$?=(-6)^2-4(2m - 1)\geqslant 0，$?
?$ 36-8m + 4\geqslant 0，$?
?$ 40-8m\geqslant 0，$?
?$ $?解得?$m\leqslant 5。$?
?$ $?假設(shè)存在實(shí)數(shù)?$m，$?滿足?$(x_{1} - 1)(x_{2} - 1)=\frac {6}{m - 5}，$?
?$ $?因?yàn)?$x_{1} + x_{2} = 6，$??$x_{1}x_{2} = 2m - 1，$?
?$ $?則?$x_{1}x_{2}-(x_{1} + x_{2})+1=\frac {6}{m - 5}，$?
?$ $?即?$2m - 1-6 + 1=\frac {6}{m - 5}，$?
?$ 2m-6=\frac {6}{m - 5}，$?
?$ (2m - 6)(m - 5)=6，$?
?$ 2\ \mathrm {m^2}-10m-6m + 30 = 6，$?
?$ 2\ \mathrm {m^2}-16m + 24 = 0，$?
?$\mathrm {m^2}-8m + 12 = 0，$?
?$ (m - 2)(m - 6)=0，$?
?$ $?解得?$m_{1} = 2，$??$m_{2} = 6。$?
?$ $?因?yàn)?$m\leqslant 5$?且?$m - 5\neq 0，$?所以?$m = 2。$?
經(jīng)檢驗(yàn)，?$m = 2$?是原分式方程的解，且符合題意。
所以假設(shè)成立，即存在實(shí)數(shù)?$m = 2，$?滿足?$(x_{1} - 1)(x_{2} - 1)=\frac {6}{m - 5}。$?
