Maybe the following argument proves (1) under the hypothesis that $\vert G \vert$ is invertible in $R$. Is it correct?
Since $A$ is finitely presented over $R$, there exist a subring $R_\alpha$ of $R$ and an $R_\alpha$-algebra $A_\alpha$ of finite type such that $R_\alpha$ is of finite type over $\mathbb{Z}$ and $A \simeq R \otimes_{R_\alpha} A_\alpha$. Consider the filtrant inductive system $\{R_\lambda \}_{\lambda \geq \alpha}$ of subrings $R_\lambda$ of $R$ which are finitely generated extension of $R_\alpha$. Set $A_\lambda = R_\lambda \otimes_{R_\alpha} A_\alpha$. We are in the situation of [EGA Lemme IV.8.8.2.1], hence the $R$-automorphisms $g_1, \dots g_n$ of $A$ come from $R_\lambda$-homomorphism $g_1^\lambda, \dots, g_n^\lambda$ of $A_\lambda$, for $\lambda \gg \alpha$. The injectivity of the homomorphism (8.8.2.2) implies that, for $\lambda \gg \alpha$, the $g_i^\lambda$ are automorphisms and satisfy the presentation of the group $G$. Besides, if $G$ acts faithfully on $A$, we can require that it acts faithfully also on $A_\lambda$, for some $\lambda \gg \alpha$. Since $\vert G \vert$ is invertible in $R$, then it is invertible in $R_\lambda$, then taking invariants commutes with base change: $A^G = (R \otimes_{R_\lambda} A_\lambda)^G \simeq R \otimes_{R_\lambda} A_\lambda^G$, then $A^G$ is of finite presentation over $R$ because $A_\lambda^G$ is of finite type over the noetherian ring $R_\lambda$.