Question 1018321
There may be a better way to get to the same answer.
Here is a cumbersome one.
 
Using the quadratic formula, we find that the solutions to {{{x^2-kx+3k-2=0}}} ,
if any, are given by
{{{x = (k +- sqrt(k^2-4*1*(3k-2)))/(2*1) = (k +- sqrt(k^2-12x+8))/2 = k/2 +- sqrt(k^2-12x+8)/2 }}} .
For the solutions to exist, we need to have
{{{k^2-12x+8>=0}}} , and only if {{{k^2-12x+8>0}}} there are two different solutions.
Since the solutions to {{{k^2-12x+8>=0}}} are
{{{k = (12 +- sqrt(12^2-4*1*8))/(2*1) = (12 +- sqrt(144-32))/2 = (12 +- sqrt(112))/2 = (12 +- 4sqrt(7))/2 =6 +- 2sqrt(7) }}} ,
two different solutions to {{{x^2-kx+3k-2=0}}} exist only if
{{{k<6-2sqrt(7) }}} or {{{k>6+2sqrt(7) }}} .
 
If {{{k>6+2sqrt(7)}}} , {{{k/2 + sqrt(k^2-12k+8)/2 >k/2=(6+2sqrt(7))/2=3+sqrt(7)>5}}} .
So, if {{{k>6+2sqrt(7)}}} , there is one solution (the one above) that is
neither within {{{-1< x <0 }}} nor within {{{1< x <2}}} . 
 
If {{{k<6-2sqrt(7)=about0.7}}} ,
and {{{x^2-kx+3k-2=0}}} is to have
one solution within {{{-1< x <0 }}} , and
another solution within {{{1< x <2}}} , it must be
{{{-1<k/2-sqrt(k^2-12k+8)/2<0 }}} , and {{{1<k/2+sqrt(k^2-12k+8)/2<2}}} .
 
{{{1<k/2+sqrt(k^2-12k+8)/2<2}}}-->{{{2<k+sqrt(k^2-12k+8)<4}}}-->{{{2-k<sqrt(k^2-12x+8)<4-k}}}
Since {{{2-k>0}}} ,
{{{2-k<sqrt(k^2-12k+8)<4-k}}}-->{{{(2-k)^2<k^2-12k+8<(4-k)^2}}}-->{{{4-4k+k^2<k^2-12k+8<16-8k+k^2}}}-->{{{4-4k<-12k+8<16-8k}}}-->{{{4+8k<8<16+4k}}}-->{{{system(8k<4,"and",-8<4k)}}}-->{{{system(k<1/2,"and",-2<k)}}}-->{{{-2<k<1/2}}}
 
{{{-1<k/2-sqrt(k^2-12x+8)/2<0}}}-->{{{-2<k-sqrt(k^2-12x+8)<0 }}}-->{{{-2-k<-sqrt(k^2-12x+8)<-k}}}-->{{{k+2>sqrt(k^2-12x+8)>k}}} .
As we already found out (above) that we need {{{k>-2}}} , {{{k+2>0}}} , so
{{{k+2>sqrt(k^2-12x+8)}}}<-->{{{(k+2)^2>k^2-12x+8}}}<-->{{{k^2+4k+4>k^2-12x+8}}}<-->{{{4k+4>-12x+8}}}<-->{{{16k>4}}}<-->{{{k>1/4}}} .
 
In sum, the answer is {{{highlight(1/4<k<1/2)}}} .