a. Any vector in the span of [tex]S[/tex] is a linear combination of the vectors in [tex]S[/tex]. The simplest one we could come up with is the addition of the two vectors we know:
[tex]p(x)=(-8x^2+4x-5)+(-2x+5)=\boxed{-8x^2+2x}[/tex]
b. Since one vector is quadratic while the other is purely linear, there is no choice of [tex]c_1,c_2[/tex] such that
[tex]c_1(-8x^2+4x-5)+c_2(-2x+5)=\boxed{x}[/tex]
because the only way to eliminate the [tex]x^2[/tex] term is to pick [tex]c_1=0[/tex], but there's no way to eliminate the remaining constant term.