Any tangent line to [tex]y=x^2-x[/tex] at a point [tex](a,b)=(a,a^2-a)[/tex] will have slope [tex]y'(a)=2x-1\bigg_{x=a}=2a-1[/tex]. Such a line has equation
[tex]y-(a^2-a)=(2a-1)(x-a)\implies y=(2a-1)x-a^2[/tex]
For this (these) line(s) to pass through (2, -3), you must have
[tex]-3=(2a-1)\times2-a^2\implies a^2-4a-1=0\implies a=2\pm\sqrt5[/tex]
So the two lines in questions are [tex]y=(2a-1)x-a^2[/tex] for these values of [tex]a[/tex].
To show there is no tangent line that passes through (2, 7), you have to show there is no solution for [tex]a[/tex] in
[tex]7=(2a-1)\times2-a^2[/tex]
You have
[tex]a^2-4a+9=0[/tex]
but this has no real solutions, so no such tangent line exists.
