Predicate Logic: Finding languages $L_i$ to fulfill well formed formula

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$S$ is an one-place relation symbol, and for $i=\{1,2,3\}$ let $R_i$ be a two place relation symbol.

$F_i: \exists x \forall y (R_i(x,y)\rightarrow S(y))$

Let $A=\{a,b\}$. For every $w\in A^+$ and for every $i$ the Interpretation $(D_w,I_w)$ with:

$D_w = \mathbb{Z}_{|w|} = \{i \in \mathbb{N}_0 | 0 \leq i < |w|\}$

$I_w(S) = \{y \in D| w(y) = a\}$

a) $I_w(R_1) =\{(x,y) \in D \times D| x\neq y)\}$

b) $I_w(R_2) =\{(x,y) \in D \times D| x\leq y)\}$

c) $I_w(R_3) =\{(x,y) \in D \times D| x+y$ is even $\}$

Since the Formula $F_i$ doesn't have free variables, $\beta$ can be chosen freely.

For every $i$ find the language $L_i \subseteq A^*$ explicitly, that has every word $w \in A^+$, for which $val_{D_w,I_w,\beta}(F_i)= true$.

a) $F_1: \exists x \forall y (R_1(x,y)\rightarrow S(y)) \\ \rightarrow \exists x \forall y ((x \neq y)\rightarrow (w(y)=a))$

b)$F_2: \exists x \forall y (R_2(x,y)\rightarrow S(y)) \\ \rightarrow \exists x \forall y ((x \leq y)\rightarrow (w(y)=a))$

c)$F_3: \exists x \forall y (R_3(x,y)\rightarrow S(y)) \\ \rightarrow \exists x \forall y ((x+y = 2k)\rightarrow (w(y)=a))$

I'm new to this topic. In a) it means that if the pair of numbers aren't the same, then the number $y$, from the domain, stands for the word $a$. Similarly for b) and c).

But how do I find the languages $L_i$? How do I go on from here?

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a) There is some $x$ such that all positions that are different from $x$ have the symbol $a$. This means that $L_1$ consists of all words that have at most one symbol, which is not $a$.

b) There is some $x$ such that all positions not to the left of $x$ have the symbol $a$. This means that $L_2$ consists of all words that have a suffix which consists only of symbols $a$ and has length at least one.

c) There is some $x$ such that if the sum $x+y$ is even, then position $y$ is $a$. If $x$ is even, then the sum is even for all even $y$; then all words where all even positions are $a$ fulfill the formula. If $x$ is odd, then the sum is even for all odd $y$; then all words where all odd positions are $a$ fulfill the formula. $L_3$ consists of these two classes of words.