I don't understand what you've been doing. Let's call $u$ and $v$ the vectors from $S$ and $w$ the one from (a). What do you want to know is if there exists real numbers $x$ and $y$ such that
$ w = xu + y v \ . $
Right?
But this is a system of linear equations, with four equations (right) and just two unknowns ($x$ and $y$). This one:
$ \begin{pmatrix} -42 \\ 113 \\ -112 \\ -60 \end{pmatrix} = x \begin{pmatrix} 6 \\ -7 \\ 8 \\ 6 \end{pmatrix} + y \begin{pmatrix} 4 \\ 6 \\ -4 \\ 1 \end{pmatrix} $
Which you can write as
$ \begin{pmatrix} 6 & 4 & \vert & -42 \\ -7 & 6 &\vert & 113 \\ 8 & -4 &\vert & -112 \\ 6 & 1 & \vert & -60 \end{pmatrix} $
If I'm not wrong, this time this system has no solution at all (and so, $w$ is NOT a linear combination of $u$ and $v$), but you don't have to think that, just because a system of linear equations has more unknowns than equations this necessarily means that it has no solutions. For instance, this one
$ \begin{align} x + y &=& 1 \\ x + y &=& 1 \\ x + y &=& 1 \end{align} $
Has two unknowns and three equations, but an infinite number of solutions. "Oh, but it's always the same equation, so it doesn't count". -Is that what you're thinking? Well, try to solve this one:
$ \begin{align} x + y &=& 1 \\ x - y &=& 0 \\ 2x &=& 1 \end{align} $