The solutions of a homogeneous system of linear differential equations of the first order $$\frac{d \vec x}{d t} = A \vec x$$ are $$\vec x = e^{ \lambda t} \vec v $$ where $ \lambda $ are the eigenvalues of A and $\vec v$ are the associated eigenvectors.
So the first and second options are true. As for the others I'm not sure.
Consider the matrix \begin{align} A =&\ \begin{pmatrix} 1 & -1 & 1 & 0\\ -1 & 2 & 0 & 1\\ 1 & 2 & 0 & 0 \\ -2 & -1 & 0 & 0 \end{pmatrix} \begin{pmatrix} -2 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 & 0\\ -1 & 2 & 0 & 1\\ 1 & 2 & 0 & 0 \\ -2 & -1 & 0 & 0 \end{pmatrix}^{-1}\\ =&\ \frac{1}{3}\begin{pmatrix} 0 & 0 & 4 & 5\\ 0 & 0 & -6 & -6\\ 0 & 0 & -2 & 2 \\ 0 & 0 &-2 & -7 \end{pmatrix} \end{align} then it's clear that $v=(-1, 2 ,2, -1)^T$ is an eigenvector with eigenvalue $-1$. Likewise, $w=(1, -1, 1, -2)^T$ is an eigenvector with eigenvalue $-2$. However, 3 is not an eigenvalue of $A$ so $(A-3I)\mathbf{x} = \mathbf{0}$ has only a trivial solution. It's also easy to check that the last statement is false.