Let $R \subseteq S\times S$ be any binary relation on a set S. Consider the sequence $R^{0}, R^{1}, R^{2}...$ defined as follows: $$R^{0} := I = \left \{ (x,x):x\in S \right \}\\ R^{n+1} :=R^{n}\cup \left( R; R^{n} \right) \quad\forall\ n\geqslant 0,$$ where $;$ means relation composition.
Suppose there exists $i\in \mathbb{N}$ such that $R^{i} = R^{i+1}$.
Q.1: Let $P(n)$ be the proposition that $\forall\ m\in \mathbb{N}$ gives $R^{n}; R^{m} = R^{n+m}$. Prove that $P(n)$ holds for all $n\in \mathbb{N}$.
Q.2: if $|S| = k$, show that $\left( R\cup R^{\leftarrow } \right)^{k^{2}}$ is an equivalence relation.
Regaring question 1, I proved that $R; I = I; R = R$ and $(R_{1}\cup R_{2});R_{3} = \left ( R_{1};R_{3} \right )\cup \left ( R_{2};R_{3} \right )$ and $R_{1};\left ( R_{2};R_{3} \right )=\left ( R_{1};R_{2} \right );R_{3}$,
but I don't know how to deal with the relationship of $R^{n}$ and $R;R^{n}$.
Regarding question 2, I am confused about the $k^{2}$, does that mean $k^{2}$ is equal to $i$? Can anyone give me some hint?
I don't think the assumption $R^{i}=R^{i+1}$ for some $i\in\mathbb{N}$ has any direct bearing on either of the two questions.
For question $1$, your wording of $P(n)$ is slightly confusing, but I am going to assume that $P(n)$ stands for the statement "For all $m$, $(R^{n};R^{m})=R^{n+m}$".
We want to proof $P(n)$ by induction on $n$. The base case $P(0)$ says "For all $m$, $(I;R^{m})=R^{m}$". You say you know how to show $(I;R)=R$, so the same kind of argument should work for any given $m$ (just treat each $R^{m}$ as an individual relation). For the induction hypothesis, assume $P(n)$ holds. We want to show $P(n+1)$ holds, which is the statement "For all $m$, $(R^{n+1};R^{m})=R^{n+m+1}$. Using the general properties of arbitrary relations $R_{1},R_{2},R_{3}$ that you say you know how to show, we can argue as follows: $$ (R^{n+1};R^{m})=((R^{n}\cup (R;R^{n}));R^{m})=(R^{n};R^{m})\cup ((R;R^{n});R^{m})=(R^{n};R^m)\cup (R;(R^{n};R^{m})) $$ Now you should be able to finish using the induction hypothesis and the definition of $R^{n+m+1}$.
For question 2, we assume $|S|=k$ and we want to show $(R\cup R^{\leftarrow})^{k^2}$ is an equivalence relation. In fact, it seems to me that we can show $(R\cup R^{\leftarrow})^{k}$ is an equivalence relation.
Here are some suggestions. To make it simpler let's let $F$ denote the relation $R\cup R^{\leftarrow}$. I would first show that $F$ is symmetric, and from that show that $F^{n}$ is symmetric and reflexive for any $n$.$^{\ast}$ So we just need to show that $F^{k}$ is transitive. Here is a strategy:
Step 1. Let $F^*=\bigcup_{n=1}^\infty F^{n}$ and show that $F^*$ is transitive. This should be very easy.
Step 2. By induction on $n\geq 1$, show that $(a,b)\in F^{n}$ if and only if there is a sequence of elements $a=x_{0},x_{1}, \ldots, x_{t-1}, x_{t}=b$ of $S$ such that $t\leq n$ and $(x_{i},x_{i+1})\in F$ for all $i$.
Step 3. We show that $F^{k}=F^*$, which tells us that $F^{k}$ is transitive. So let $(a,b)\in F^*$. We can choose the smallest $n$ such that $(a,b)\in F^{n}$. If $n\leq k$ then we are done so suppose $n>k$. Let $a=x_{0}, x_{1}, \ldots, x_{n}=b$ be as in step 2. Among the $n$ elements $x_{0}, \ldots, x_{n-1}$ there must be a repetition since $|S|=k$. So there is $0\leq i<j<n$ with $x_{i}=x_{j}$. Now remove $x_{i+1},\ldots, x_{j}$ from the sequence and we have a shorter path for $(a,b)$. This contradicts the minimality assumption on $n$.
$^{\ast}$ A word of caution about this notation. If $R$ is a relation then $R^{1}=I\cup (R;I)=I\cup R$. So $R^{1}$ might not be the same as $R$. Indeed, $R^{1}=R$ if and only if $R$ is reflexive.