Define an isomorphism from $\mathbf{R}^2 $to $S$, where $S=\operatorname{span} .... $(image) Not a clue what to do with this question.  We only briefly touched on isomorphisms in class,and I can't find any example online that uses transformations on two basis vectors, as in this question.  Help a plebe out? 
 A: For part $a)$ You can define a Linear map $T:\mathbf{R^2}\to S$ as follows
$$T\left(c_1\begin{pmatrix}1\\0\\\end{pmatrix}+c_2\begin{pmatrix}0\\1\\\end{pmatrix}\right) = c_1\begin{pmatrix}3\\1\\4\\\end{pmatrix}+c_2\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$$
The remaining task is then to show that with the above definition $T$ is an isomorphism. You can show this by proving that $\operatorname{null}T = \{0\}$ and $\operatorname{range}T =S$. The first of these two claims follows from the linear independence of $\begin{pmatrix}3\\1\\4\\\end{pmatrix}$ and $\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$ and the latter claim is evident if you consider that $S$ by definition is the span of these two vectors.
For $b)$ Note that $v = \begin{pmatrix}10\\11\\15\\\end{pmatrix} = 4\cdot\begin{pmatrix}3\\1\\4\\\end{pmatrix}-1\cdot\begin{pmatrix}2\\-7\\1\\\end{pmatrix}$ consequently with the above choice of $T$ 
$$T^{-1}(v) = 4\cdot\begin{pmatrix}1\\0\end{pmatrix}-1\cdot\begin{pmatrix}0\\1\end{pmatrix}= \begin{pmatrix}4\\-1\end{pmatrix}$$ 
