Let $A=\begin{pmatrix} 5 & -2 & -7 \\ -4 & 3 & 4 \\ 6 & -3 & -8 \end{pmatrix} $ and $\vec{v}=\left(\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right)$.
The first part of the problem asked to calculate $(A-I)\vec{v}$ and $(A-I)^2\vec{v}$, which I calculated to be $\left(\begin{array}{c} -1 \\ -2 \\ 0 \end{array}\right)$ and $\vec{0}$ respectively.
The question then asks to solve the initial value problem $\vec{y'}=A\vec{y}$ with $\vec{y}(0)=\vec{v}$.
I don't understand how to use what we calculated in the earlier part to help to solve this initial value problem. I determined from first principles that the eigenvalues are $\lambda=1,1,-2$ and I know that the solution of our initial value problem will take the form $\vec{y}=e^{tA}\vec{v}\ $, but I don't quite understand the significance of $(A-I)\vec{v}$ and $(A-I)^2\vec{v}$.
The solution to your IVP can be formulated in terms of the exponential of the matrix $A$ as $y(t) = \exp(At) v$. Now, let's rewrite this as $y(t) = \exp(It + (A-I)t)v = \exp(It)\exp((A-I)t)v$, where $I$ denotes unit matrix. The first term $\exp(It)$ is just a $\exp(t)$ multiple of the unit matrix. As for the second term:
$$ \exp((A-I)t)v = (I + (A-I)t + \frac{1}{2}(A-I)^2t^2 + \ldots\,)\,v $$
Since $(A-I)^2v$ yields zero vector, all the higher terms in the matrix exponential will be zero as well.
Thus in total:
$$ y(t) = \exp(t)\bigl( v + t\,(A-I)v + \frac{t^2}{2}\,(A-I)^2v\bigr)$$