jmc

We use the fact that the sum and product of the roots of a quadratic equation $a{x}^2+bx+c=0$ are given by $-b/a$ and $c/a,$ respectively. This means that $a+b=7/2$ and $ab=2/2=1.$ Now we manipulate the expression $\frac{1}{a-1}+\frac{1}{b-1}$ to get: $$\frac{1}{a-1}+\frac{1}{b-1}=\frac{b-1}{(a-1)(b-1)}+\frac{a-1}{(a-1)(b-1)}=\frac{(a+b)-2}{(a-1)(b-1)}.$$ But the denominator $$(a-1)(b-1)=ab-a-b+1=(ab)-(a+b)+1=1-7/2+1=2-7/2,$$ whereas the numerator $a+b-2=7/2-2.$

Thus, our answer is $\frac{7/2-2}{2-7/2}=\boxed{-1}.$