First question: no: consider e.g. $G:=\left[\begin{matrix} 0 & 1 \\0 & 0  \end{matrix}\right]$ and $G_0:=\left[\begin{matrix} 1 & 1 \\0 & 0  \end{matrix}\right]$ on $\mathbb{R}^2$. 

Second question: no: consider e.g. any  symmetric compact operator with finite dimensional kernel on an infinite dimensional Hilbert space. Then $G_0$ is still compact, so $0\in\sigma(G_0)$.