We will use that $\mathcal{O}_{\mathbf{P}^n_R}(1)$ is an ample invertible sheaf on the scheme $\mathbf{P}^n_R$. This follows directly from the definition since $\mathbf{P}^n_R$ covered by the standard affine opens $D_{+}(T_i)$. Hence by Properties, Proposition \ref{properties-proposition-characterize-ample} every finite type quasi-coherent $\mathcal{O}_{\mathbf{P}^n_R}$-module is a quotient of a finite direct sum of tensor powers of $\mathcal{O}_{\mathbf{P}^n_R}(1)$. On the other hand coherent sheaves and finite type quasi-coherent sheaves are the same thing on projective space over $R$ by Lemma \ref{lemma-coherent-Noetherian}. Thus we see (1). \medskip\noindent Projective $n$-space $\mathbf{P}^n_R$ is covered by $n + 1$ affines, namely the standard opens $D_{+}(T_i)$, $i = 0, \ldots, n$, see Constructions, Lemma \ref{constructions-lemma-standard-covering-projective-space}. Hence we see that for any quasi-coherent sheaf $\mathcal{F}$ on $\mathbf{P}^n_R$ we have $H^i(\mathbf{P}^n_R, \mathcal{F}) = 0$ for $i \geq n + 1$, see Lemma \ref{lemma-vanishing-nr-affines}. Hence (2) holds. \medskip\noindent Let us prove (3) and (4) simultaneously for all coherent sheaves on $\mathbf{P}^n_R$ by descending induction on $i$. Clearly the result holds for $i \geq n + 1$ by (2). Suppose we know the result for $i + 1$ and we want to show the result for $i$. (If $i = 0$, then part (4) is vacuous.) Let $\mathcal{F}$ be a coherent sheaf on $\mathbf{P}^n_R$. Choose a surjection as in (1) and denote $\mathcal{G}$ the kernel so that we have a short exact sequence $$ 0 \to \mathcal{G} \to \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j) \to \mathcal{F} \to 0 $$ By Lemma \ref{lemma-coherent-abelian-Noetherian} we see that $\mathcal{G}$ is coherent. The long exact cohomology sequence gives an exact sequence $$ H^i(\mathbf{P}^n_R, \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j)) \to H^i(\mathbf{P}^n_R, \mathcal{F}) \to H^{i + 1}(\mathbf{P}^n_R, \mathcal{G}). $$ By induction assumption the right $R$-module is finite and by Lemma \ref{lemma-cohomology-projective-space-over-ring} the left $R$-module is finite. Since $R$ is Noetherian it follows immediately that $H^i(\mathbf{P}^n_R, \mathcal{F})$ is a finite $R$-module. This proves the induction step for assertion (3). Since $\mathcal{O}_{\mathbf{P}^n_R}(d)$ is invertible we see that twisting on $\mathbf{P}^n_R$ is an exact functor (since you get it by tensoring with an invertible sheaf, see Constructions, Definition \ref{constructions-definition-twist}). This means that for all $d \in \mathbf{Z}$ the sequence $$ 0 \to \mathcal{G}(d) \to \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j + d) \to \mathcal{F}(d) \to 0 $$ is short exact. The resulting cohomology sequence is $$ H^i(\mathbf{P}^n_R, \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j + d)) \to H^i(\mathbf{P}^n_R, \mathcal{F}(d)) \to H^{i + 1}(\mathbf{P}^n_R, \mathcal{G}(d)). $$ By induction assumption we see the module on the right is zero for $d \gg 0$ and by the computation in Lemma \ref{lemma-cohomology-projective-space-over-ring} the module on the left is zero as soon as $d \geq -\min\{d_j\}$ and $i \geq 1$. Hence the induction step for assertion (4). This concludes the proof of (3) and (4). \medskip\noindent In order to prove (5) note that for all sufficiently large $d$ the map $$ H^0(\mathbf{P}^n_R, \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j + d)) \to H^0(\mathbf{P}^n_R, \mathcal{F}(d)) $$ is surjective by the vanishing of $H^1(\mathbf{P}^n_R, \mathcal{G}(d))$ we just proved. In other words, the module $$ M_k = \bigoplus\nolimits_{d \geq k} H^0(\mathbf{P}^n_R, \mathcal{F}(d)) $$ is for $k$ large enough a quotient of the corresponding module $$ N_k = \bigoplus\nolimits_{d \geq k} H^0(\mathbf{P}^n_R, \bigoplus\nolimits_{j = 1, \ldots, r} \mathcal{O}_{\mathbf{P}^n_R}(d_j + d) ) $$ When $k$ is sufficiently small (e.g.\ $k < -d_j$ for all $j$) then $$ N_k = \bigoplus\nolimits_{j = 1, \ldots, r} R[T_0, \ldots, T_n](d_j) $$ by our computations in Section \ref{section-cohomology-projective-space}. In particular it is finitely generated. Suppose $k \in \mathbf{Z}$ is arbitrary. Choose $k_{-} \ll k \ll k_{+}$. Consider the diagram $$ \xymatrix{ N_{k_{-}} & N_{k_{+}} \ar[d] \ar[l] \\ M_k & M_{k_{+}} \ar[l] } $$ where the vertical arrow is the surjective map above and the horizontal arrows are the obvious inclusion maps. By what was said above we see that $N_{k_{-}}$ is a finitely generated $R[T_0, \ldots, T_n]$-module. Hence $N_{k_{+}}$ is a finitely generated $R[T_0, \ldots, T_n]$-module because it is a submodule of a finitely generated module and the ring $R[T_0, \ldots, T_n]$ is Noetherian. Since the vertical arrow is surjective we conclude that $M_{k_{+}}$ is a finitely generated $R[T_0, \ldots, T_n]$-module. The quotient $M_k/M_{k_{+}}$ is finite as an $R$-module since it is a finite direct sum of the finite $R$-modules $H^0(\mathbf{P}^n_R, \mathcal{F}(d))$ for $k \leq d < k_{+}$. Note that we use part (3) for $i = 0$ here. Hence $M_k/M_{k_{+}}$ is a fortiori a finite $R[T_0, \ldots, T_n]$-module. In other words, we have sandwiched $M_k$ between two finite $R[T_0, \ldots, T_n]$-modules and we win.