The answer is No. Lemma 1. Let $n$ be a positive integer. Let $x_1,\ldots,x_n$ be pairwise distinct nonnegative real numbers and let $y_1,\ldots,y_n$ be real numbers. Then there exists a nonconstant even polynomial $f(x)\in\mathbb R[x]$ such that $f(x_i)=y_i$ for all $i\in\{1,2,\ldots,n\}$. Proof. Since $x_1,\ldots,x_n$ are nonnegative and pairwise distinct, then so are $x_1^2,\ldots,x_n^2$. By Lagrange's interpolation formula, there exists a polynomial $g(x)\in\mathbb R[x]$ such that $g(x_i^2)=y_i$ for every $i\in\{1,2,\ldots,n\}$. Replacing $g(x)$ by $g(x)+\prod_{i=1}^n (x-x_i^2)$ if necessary, we can assume $g(x)$ is nonconstant. Now choose $f(x)=g(x^2)$. $\square$ Lemma 2. There exists a sequence of even polynomials $\{f_n(x)\}_{n\in\mathbb{Z}_{\ge 0}}$ such that: $f_n(x)$ is the zero polynomial if and only if $n=0$, and for every nonnegative integers $n,k$ we have $f_n(k)=f_{|n-k|}(k)$. Proof. We define $f_0(x),f_1(x),\ldots$ inductively, in this order. Initially define $f_0(x)\equiv 0$. Now, let $n$ be a positive integer and assume $f_0(x),\ldots,f_{n-1}(x)$ have been defined. By Lemma 1, there exists a nonconstant even polynomial $f_n(x)\in\mathbb R[x]$ such that $f_n(k)=f_{|n-k|}(k)$ whenever $k\in\mathbb Z$ and $1\le k\le 2n-1$. Note that whenever $1\le k\le 2n-1$ then $0\le |n-k|\le n-1$, so the definition is well-defined. It remains to prove that $f_n(k)=f_{|n-k|}(k)$ for all nonnegative integers $n,k$. We divide into cases. Case 1. $k=0$ or $k=2n$. In those cases $|n-k|=n$ so the equation $f_n(k)=f_{|n-k|}(k)$ is tautological. Case 2. $1\le k\le 2n-1$. The inequality $1\le 2n-1$ implies that $n\ge 1$, so we are done by construction. Case 3. $k\ge 2n+1$. In this case $1\le k\le 2(k-n)-1$, so $$f_{|n-k|}(k)=f_{k-n}(k)=f_{|(k-n)-k|}(k)=f_n(k).$$ Since we have exhausted all cases, this proves the lemma. $\square$ Now choose even polynomials $\{f_n(x)\}_{n\in\mathbb Z_{\ge 0}}$ in $\mathbb R[x]$ which satisfy the conditions of Lemma 2. We now extend the sequence $\{f_n(x)\}_{n\in\mathbb Z_{\ge 0}}$ to $\mathbb Z$ by defining $f_n(x)=f_{-n}(x)$ whenever $n\in\mathbb Z$ and $n<0$. Now since $f_0(x)\equiv 0$ and $f_1(x)\not\equiv 0$, then not all $f_n$ are equal. It remains to prove that $f_n(k)=f_{n+k}(k)$ for all integers $n,k$. Since $f_n(x)=f_{-n}(x)$ and $\{f_n(x)\}$ are all even polynomials, changing signs to both $n,k$ does not change the equation. Hence it is enough to assume that $n\ge 0$. We now split into cases. Case 1. $k\ge 0$. Then $$f_{n+k}(k)=f_{|(n+k)-k|}(k)=f_n(k).$$ Case 2. $k<0$. Let $a=-k$. Then $a,n\ge 0$ so $$f_n(k)=f_n(-a)=f_n(a)=f_{|n-a|}(a)=f_{|n+k|}(-k)=f_{n+k}(k).$$ This ends our proof.

The answer is Yes. Suppose $\{f_n(x)\}_{n\in\mathbb Z}$ is a series of polynomials in $\mathbb Z[x]$ satisfying \[f_n(k) = f_{n+k}(k) \quad \forall n, k \in \mathbb{Z}.\]We can substract $f_0(x)$ from all polynomials so that we can assume $f_0(x)\equiv 0$. Our goal is to prove that $f_a(x)\equiv 0$ for all $a\in\mathbb Z$. Observe that for every integer $a$ we have $f_a(a)=f_0(a)=0$. Moreover, for every integers $a,b$ we have $$a-b\mid f_a(a)-f_a(b)=-f_a(b).$$Therefore, for every integers $a,b,n$ we have $$a+b(n-1)\mid f_{a+bn}(b)=f_a(b).$$Assuming $b\ne 0$ and taking $n$ to be large enough we obtain $f_a(b)=0$ for all integers $a,b$ with $b\ne 0$. It follows that for all $a\in\mathbb Z$ the polynomial $f_a(x)$ has infinitely many roots, so $f_a(x)\equiv 0$.

a) The answer is no. Let $(f_j)$'s be polynomials in $\mathbb{Q}[x]$. We construct our sequence inductively: We first construct $f_0(x)=0$ and $f_1(x)=f_{-1}(x)=x^2-1$ Suppose we are given $f_{-n}(x), \cdots, f_n(x)$. We wish to construct $f_{n+1}(x)$ and $f_{-(n+1)}(x)$. This is easy because we know that all integers $x$ with $|x|\le 2n+1$ satisfy $f_{n+1}(x) = f_{x_j}(x)$ for some $x_j \in \{-n,\cdots,n\}$, and by Lagrange Interpolation I can construct $f_{n+1}(x)$ with these value restrictions. Same with $f_{-(n+1)}(x)$ (Explicitly, let $a_1=-(2n+1), \cdots, a_{2n+1}=-1, a_{2n+2} = 1, \cdots, a_{4n+2}=2n+1$, then $$f_{n+1}(x) = \sum_{j=1}^{4n+2} f_{n + 1 - |a_j|}(a_j) \prod\limits_{k\ne j, 1\le k\le 4n+2} \frac{x-a_k}{a_j-a_k}$$ b) The answer is yes. Assume for the sake of contradiction that $f_j(x) \ne f_k(x)$. Let $S = \{ s \mid f_j(s)=f_k(s)\}$ which is a finite set. We pick $t\notin S, t\ne 0$ and $p>|t|$ a prime such that $p\nmid f_j(t)-f_k(t)$. Then by Bezout since $\gcd(p,t)=1$ there exists $l$ such that $t\mid k+l-j$ and $t+p \mid l$ (which is doable since $\gcd(t,t+p)=\gcd(t,p)=1$). Then $$f_k(t)\equiv f_k(t+p) = f_{k+l}(t+p) \equiv f_{k+l}(t) = f_j(t) (\bmod\; p)$$ Contradiction

Nice solution, though a small verification should be done in (a), namely the identity for all $n,k$. Also, in part (b) you should also require $t\ne 0$, but nothing else changes.

The answer is Yes. First, we prove a lemma. Lemma. Let $m,n,a\in\mathbb{Z}$. Then $$m-n\mid f_m(a)-f_n(a).$$Proof. This follows from a chain of equivalences \begin{align*} f_m(a)&\equiv f_m(a+n-m)\\ &=f_{a+n}(a+n-m)\\ &\equiv f_{a+n}(a)\\ &=f_n(a)\quad\pmod{m-n} \end{align*}as desired. $\square$ Note that by induction $f_n(k)=f_{n+mk}(k)$ for all integers $n,k,m$. Now fix integers $m,n$. Then by the Lemma above, for all $k,a\in\mathbb Z$ we have $$m+ka-n\mid f_{m+ka}(a)-f_n(a)=f_m(a)-f_n(a).$$In the case $a\ne 0$, taking $k\to\infty$ implies that $f_m(a)=f_n(a)$. Therefore, the polynomial $f_m(x)-f_n(x)$ has infinitely many roots so $f_m(x)=f_n(x)$.

Answer: No We construct the $f_i$'s inductively. Set $f_0\equiv 0$. Given the set of constructed polynomials $\{f_{-a},f_{-a+1},\ldots, f_b\}$ we can construct $f_{-a-1}$ or $f_b$ (WLOG $f_b$) such that $f_b(i)=f_{b-i}(i)$ for every $i=1,2,\ldots, a+b$, and $f_b$ exists by Lagrange's interpolation. Now, FTSOC suppose that at some point in this process, there exists $n,k$ such that $f_n(k)\neq f_{n+k}(k)$. Assume WLOG $f_n$ was constructed first. Then, $f_{n+k}(k)=f_{n+k-k}(k)=f_n(k)$ by construction, a contradiction. Thus, this works.

Answer: Yes Assume WLOG $f_0$ is the zero polynomial by subtracting all polynomials by $f_0(x)$. Thus, $$f_{kn}(n)=f_{(k-1)n}(n)=\ldots f_0(n)=0$$However, for integers $a,b$ we have that $$f_a(n)\equiv f_a(n+b-a)\equiv f_{n+b}(n+b-a)\equiv f_{b}(n+b-a)\equiv f_b(n)\pmod{a-b}$$so if $f_b(m)\neq 0$ for some $b,m\in\mathbb Z$, we have $$km-b\mid f_{km}(m)-f_b(m)=-f_b(m)$$But we take $k$ to be arbitrarily large, implying $f_b(m)$, a contradiction. Thus, $f_b(m)=0$ for all $b,m\in\mathbb Z$ so $f_n\equiv f_m$ for all $m,n$.