for your first question, since $\displaystyle \lambda$ is not an eigenvalue of A, the matrix $\displaystyle \lambda I -A$ is invertible. let $\displaystyle \bold{u}=(\lambda I - A)^{-1} \bold{v},$
and $\displaystyle \bold{x}(t)=e^{\lambda t}\bold{u}.$ then $\displaystyle \frac{d \bold{x}}{dt}=\lambda e^{\lambda t}\bold{u}=\lambda \bold{x},$ and hence: $\displaystyle \frac{d \bold{x}}{dt} - A \bold{x} = (\lambda I - A)\bold{x}=e^{\lambda t}(\lambda I - A)\bold{u}=e^{\lambda t}\bold{v},$ which is
what you were asked to prove. now, the example in your first question should be very easy for you to do!
for the second question, call the first differential equation (1), the second (2) and the third (3). first differentiate
both sides of (1) with respect to $\displaystyle t$ to get: $\displaystyle x'' = 5x' + 3y' + 2e^{2t}.$ call this (4). now plug $\displaystyle y'$ from (2) into (4) to get
$\displaystyle x'' - 5x' - 6x = 6e^t + 2e^{2t},$ which is a simple linear equation and gives you the general solution for $\displaystyle x.$ now that we
have $\displaystyle x,$ we can easily find $\displaystyle y$ and $\displaystyle z,$ because by (1): $\displaystyle y=\frac{x'-5x-e^{2t}}{3},$ and by (3): $\displaystyle z=\int(x + y + e^t)dt. \ \ \ \square$