The multiplicity of a root $r$ of a polynomial $p(x)$ is the number of times that $x-r$ appears as a factor when $p(x)$ is fully factored into linear factors. You have the equation $\lambda^2(\lambda-1)=0$, which is fully factored into the linear factors $\lambda$, $\lambda$, and $\lambda-1$. Thus, $0$ is the root of multiplicity $2$, and $1$ is the root of multiplicity $1$.
Now you want to find the eigenvectors. For a given eigenvalue $\lambda$, these are the vectors $v$ such that $Av=\lambda v$ or, equivalently, $(A-\lambda I)v=\vec 0$. For $\lambda=0$ this equation becomes
a matrix equation that you solve by your favorite method. I’d just row-reduce the augmented matrix to
and choose a basis for the two-dimensional solution space.
Then do the same thing for $\lambda=1$; your starting point should be the equation
This will actually be a little easier, since the solution space will be one-dimensional.