I have explained the two questions and then showed my effort on question $(1)$ as follows (Please at least check my effort below and suggest to make it perfect):
Let $R, S,T$ be three commutative rings with $R \subset S \subset T$ and the map $ S \to T$ is faithfully flat.
Assumme a functor $F: \text{SMod} \to \text{TMod}$ between two categories, where $\text{SMod}$ is the category of $S$ modules and $\text{TMod}$ is the category of $T$-modules. (We can assume more suitable conditions if needed)
Now consider a finite flat $R$-alebra $A$.
Questions:
$(1)$ I want to show the functor $ \text{SMod} \otimes A \to \text{TMod} \otimes A$ is fully faithful provided the functor $F: \text{SMod} \to \text{TMod}$ is fully faithful.
$(2)$ I want to show if the functor $F: \text{SMod} \to \text{TMod}$ is equivalence then it induces an equivalence functor $\text{SMod} \otimes A \to \text{TMod} \otimes A$.
Note 1: here by notation $\text{SMod} \otimes A$, I want to mean we are taking action of the finite flat $R$-algebra $A$ (or scalar extension by $A$)
Note 2: You can improve my question by adding more conditions if necessary in order to answer question (1).
$$------------------------------------$$ My efforts:
Let for brevity, $\mathscr{C}=\text{SMod}$ and $\mathscr{D}=\text{TMod}$.
$(1)$ Given that $F: \mathscr{C} \to \mathscr{D}$ is fully faithful. So for any $S$-modules $M,N \in \mathscr{C}$, the functor $F$ induces the map $$F_{M,N}: \text{Hom}_{\mathscr{C}}(M,N) \to \text{Hom}_{\mathscr{D}}(F(M),F(N)),$$ which is bijective. So for any $S$-module homomorphism $f: M\to N$ in category $\mathscr{C}$, we get a unique $T$-module homomorphism $F_{M,N}(f): F(M) \to F(N)$ in category $\mathscr{D}$ by the injectiveness property. Also by the surjective property, for every $T$-module homomorphism $F_{M,N}(g): F(M) \to F(N)$ in category $\mathscr{D}$ there exists a $S$-module homomorphism $f: M \to N$ in category $\mathscr{C}$ which corresponds to each other.
Now we consider the action of the finite flat $R$-algebra $A$ on every modules of $\mathscr{C}$ and $\mathscr{D}$ such as $ M \otimes_R A$ or $N \otimes_R A$ or $F(M) \otimes_R A$ or $F(N) \otimes_R A$ etc. We can then think of the functor $F: M\to N$ and the $S$-module homomorphism $f:M \to N$ as follows: $$ {\color{red}{F \otimes 1}}: \mathscr{C} \otimes A \to \mathscr{D} \otimes A, \ \text{and} \ f \otimes 1: M \otimes_R A \to N \otimes_R A,$$ (where $A$ can be thought of as a category of single object (a $R$-module) along identity map $1$).
Thus it seems, by the property of product categories, $${\color{red}{(F \otimes 1)}}_{M,N} (f \otimes 1)=F_{M,N}(f) \otimes F_{M,N}(1),$$ where $1$ denotes the identity map. So it seems that the functor $F \otimes A$ induces the bijective map \begin{align} {\color{red}{(F \otimes 1)}}_{M,N}: &\text{Hom}_{C \otimes A}(M \otimes_R A,N \otimes_R A) \\ & \to \text{Hom}_{D \otimes A}({\color{red}{(F \otimes 1)}}(M \otimes_R A), {\color{red}{(F \otimes 1)}}(N \otimes_RA)), \end{align} which answer question (1), I think. Perhaps we can consider the following diagram for better understanding: \begin{align} \matrix{ \text{SMod} & \overset{F}{ \longrightarrow} & \text{TMod} \cr \downarrow && \downarrow \cr \text{SMod} \otimes A & \overset{{\color{red}{F \otimes 1}}}{\longrightarrow} & \text{TMod} \otimes A} \end{align}
Any comment on this approach.
Edit: The flatness property of $A$ and faithful flatness of the map $S \to T$ is enough to get the above bijective map to my intuition