Assume $R_0 \to A_0$ syntomic, in particular a local complete intersection (More on Algebra, Lemma \ref{more-algebra-lemma-syntomic-lci}). Choose a presentation $A_0 = R_0[x_1, \ldots, x_n]/J_0$. Set $C_0 = \text{Sym}^*_{A_0}(J_0/J_0^2)$. Note that $J_0/J_0^2$ is a finite projective $A_0$-module (Algebra, Lemma \ref{algebra-lemma-syntomic-presentation-ideal-mod-squares}). By Lemma \ref{lemma-improve-presentation} the ring map $A_0 \to C_0$ is smooth and we can find a presentation $C_0 = R_0[y_1, \ldots, y_m]/K_0$ with $K_0/K_0^2$ free over $C_0$. By Algebra, Lemma \ref{algebra-lemma-huber} we can assume $C_0 = R_0[y_1, \ldots, y_m]/(\overline{f}_1, \ldots, \overline{f}_c)$ where $\overline{f}_1, \ldots, \overline{f}_c$ maps to a basis of $K_0/K_0^2$ over $C_0$. Choose $f_1, \ldots, f_c \in R[y_1, \ldots, y_c]$ lifting $\overline{f}_1, \ldots, \overline{f}_c$ and set $$ C = R[y_1, \ldots, y_m]/(f_1, \ldots, f_c) $$ By construction $C_0 = C/IC$. By Algebra, Lemma \ref{algebra-lemma-localize-relative-complete-intersection} we can after replacing $C$ by $C_g$ assume that $C$ is a relative global complete intersection over $R$. We conclude that there exists a finite projective $A_0$-module $P_0$ such that $C_0 = \text{Sym}^*_{A_0}(P_0)$ is isomorphic to $C/IC$ for some syntomic $R$-algebra $C$. \medskip\noindent Choose an integer $n$ and a direct sum decomposition $A_0^{\oplus n} = P_0 \oplus Q_0$. By More on Algebra, Lemma \ref{more-algebra-lemma-lift-projective-module} we can find an \'etale ring map $C \to C'$ which induces an isomorphism $C/IC \to C'/IC'$ and a finite projective $C'$-module $Q$ such that $Q/IQ$ is isomorphic to $Q_0 \otimes_{A_0} C/IC$. Then $D = \text{Sym}_{C'}^*(Q)$ is a smooth $C'$-algebra (see More on Algebra, Lemma \ref{more-algebra-lemma-symmetric-algebra-smooth}). Picture $$ \xymatrix{ R \ar[d] \ar[rr] & & C \ar[r] \ar[d] & C' \ar[r] \ar[d] & D \ar[d] \\ R/I \ar[r] & A_0 \ar[r] & C/IC \ar[r]^{\cong} & C'/IC' \ar[r] & D/ID } $$ Observe that our choice of $Q$ gives \begin{align*} D/ID & = \text{Sym}_{C/IC}^*(Q_0 \otimes_{A_0} C/IC) \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes_{A_0} C/IC \\ & = \text{Sym}_{A_0}^*(Q_0) \otimes_{A_0} \text{Sym}_{A_0}^*(P_0) \\ & = \text{Sym}_{A_0}^*(Q_0 \oplus P_0) \\ & = \text{Sym}_{A_0}^*(A_0^{\oplus n}) \\ & = A_0[x_1, \ldots, x_n] \end{align*} Choose $f_1, \ldots, f_n \in D$ which map to $x_1, \ldots, x_n$ in $D/ID = A_0[x_1, \ldots, x_n]$. Set $A = D/(f_1, \ldots, f_n)$. Note that $A_0 = A/IA$. We claim that $R \to A$ is syntomic in a neighbourhood of $V(IA)$. If the claim is true, then we can find a $f \in A$ mapping to $1 \in A_0$ such that $A_f$ is syntomic over $R$ and the proof of (1) is finished. \medskip\noindent Proof of the claim. Observe that $R \to D$ is syntomic as a composition of the syntomic ring map $R \to C$, the \'etale ring map $C \to C'$ and the smooth ring map $C' \to D$ (Algebra, Lemmas \ref{algebra-lemma-composition-syntomic} and \ref{algebra-lemma-smooth-syntomic}). The question is local on $\Spec(D)$, hence we may assume that $D$ is a relative global complete intersection (Algebra, Lemma \ref{algebra-lemma-syntomic}). Say $D = R[y_1, \ldots, y_m]/(g_1, \ldots, g_s)$. Let $f'_1, \ldots, f'_n \in R[y_1, \ldots, y_m]$ be lifts of $f_1, \ldots, f_n$. Then we can apply Algebra, Lemma \ref{algebra-lemma-localize-relative-complete-intersection} to get the claim. \medskip\noindent Proof of (2). Since a smooth ring map is syntomic, we can find a syntomic ring map $R \to A$ such that $A_0 = A/IA$. By assumption the fibres of $R \to A$ are smooth over primes in $V(I)$ hence $R \to A$ is smooth in an open neighbourhood of $V(IA)$ (Algebra, Lemma \ref{algebra-lemma-flat-fibre-smooth}). Thus we can replace $A$ by a localization to obtain the result we want.