= Solution
The <Hilbert basis theorem> says that if $B$ is a commutative <Noetherian ring>, then $B[t]$ is <Noetherian>; iteration treats any finite number of variables. Here is a complete leading-coefficient proof. For an <ideal> $I\subseteq B[t]$, let $L_d$ be the set of degree-$d$ coefficients of its polynomials of degree at most $d$, including zero coefficients. These are <ideals> of $B$, and multiplication by $t$ gives $L_d\subseteq L_{d+1}$. The <ascending chain condition> makes the chain stationary from some $N$ onward. Choose finite generators for $L_0,\ldots,L_N$ and lift them to polynomials $f_{dj}\in I$ of degree at most $d$. If $f\in I$ has degree $m$, set $d=\min(m,N)$ and express its leading coefficient as $\sum_j a_{dj}b_j$, where $a_{dj}$ is the chosen degree-$d$ coefficient. Then $f-\sum_j b_jt^{m-d}f_{dj}$ has smaller degree. Induction proves that the finitely many lifts generate $I$. This proves the theorem, including coefficient <rings> with <zero divisors>.
Put $A=\mathbb CG$ and $L=\mathbb C[X,X^{-1}]$. The <infinite dihedral group> has normal forms $X^m$ and $X^mY$. The relations move every $Y$ to the right and reduce its exponent modulo two; the action $X:n\mapsto n+1$, $Y:n\mapsto-n$ on $\mathbb Z$ shows that these forms are distinct. Thus
$$
A=L\oplus LY,\qquad Yf(X)=f(X^{-1})Y.
$$
The <Hilbert basis theorem> makes $\mathbb C[X]$ <Noetherian>, and its <localization> $L$ is <Noetherian>: an <ideal> of a <localization> is generated by the images of the finite generators of its contraction. Since $A$ is free of rank two over $L$ on the left, it is a <Noetherian module> over $L$. Every <left ideal> of $A$ is an $L$-<submodule>, so all its ascending chains stabilize. Therefore \b[$\mathbb CG$ is left Noetherian].
For the <center of the infinite dihedral group algebra>, write an element as $f(X)+g(X)Y$. Commuting it with $X$ gives $g(X)(X^{-1}-X)=0$ in the Laurent <integral domain>, forcing $g=0$. Commuting $f(X)$ with $Y$ then gives $f(X)=f(X^{-1})$. A symmetric <Laurent polynomial> is a linear combination of $1$ and $X^j+X^{-j}$, $j\ge1$. With $z=X+X^{-1}$, the recursion
$$
S_0=2,\quad S_1=z,\quad S_{j+1}=zS_j-S_{j-1}
$$
expresses $S_j=X^j+X^{-j}$ as a polynomial in $z$. Conversely every polynomial in $z$ is central. The highest Laurent exponent of a nonzero polynomial in $z$ shows that $z$ is transcendental. Hence
$$
\boxed{R=Z(A)=\mathbb C[z],\qquad z=X+X^{-1}.}
$$
The identities $X^{-1}=z-X$ and $X^2-zX+1=0$ give $L=R+RX$. Independence follows by applying $X\mapsto X^{-1}$ to a relation $f(z)+Xg(z)=0$ and subtracting: $(X-X^{-1})g(z)=0$, so $g=f=0$. Consequently
$$
\boxed{A=R\oplus RX\oplus RY\oplus RXY}
$$
is a free $R$-<module> of rank four.
Let $M$ be a nonzero <simple module> on the left. For any $m\ne0$, $M=Am$, so it is finite over $R$. Every central $r\notin P=\operatorname{Ann}_R(M)$ acts bijectively: its kernel and image are $A$-<submodules>, and simplicity forces kernel zero and image $M$. To prove that $P$ is maximal, not merely prime, put $B=R/P$; then $M$ is a faithful finite $B$-<module>. For $0\ne b\in B$, $bM=M$. If $m_1,\ldots,m_d$ generate $M$, write $m_i=b\sum_jc_{ij}m_j$. The <adjugate matrix> identity for $I-bC$ shows that $\det(I-bC)$ annihilates every generator and hence $M$. Faithfulness makes this determinant zero in $B$. Its constant term is one, so it has form $1-bq$, proving $bq=1$. Thus $B$ is a <field>, and \b[$P$ is a <maximal ideal> of $R$]. This proves the general principle for <simple modules over an algebra finite over its center> in this instance.
Every maximal <ideal> of $\mathbb C[z]$ is $(z-\lambda)$ for some $\lambda\in\mathbb C$, by factorization of complex polynomials. The <module> $M=Am$ is therefore a quotient of the four-dimensional <algebra> $A/(z-\lambda)A$. In particular \b[$\dim_{\mathbb C}M\le4<\infty$], as required.
The actual simple dimensions are at most two. If $\lambda\ne\pm2$, choose $u\ne\pm1$ with $u+u^{-1}=\lambda$. The matrices $X=\operatorname{diag}(u,u^{-1})$ and $Y=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ give a representation of $A/(z-\lambda)A$ onto $M_2(\mathbb C)$: polynomial functions of $X$ give the two diagonal matrix units, and multiplication by $Y$ gives the off-diagonal ones. Both <algebras> have dimension four, so this is an isomorphism. If $\lambda=\pm2$, the element $X\mp1$ has square zero and is conjugated to its negative by $Y$; its generated two-sided <ideal> is square zero. It annihilates every <simple module>, since its action on a <simple module> is either zero or the whole <module>, and the latter contradicts its square being zero. The remaining quotient is $\mathbb C[C_2]$, whose two <simple modules> have $Y=\pm1$. Thus the generic <simple modules> have dimension two, and the four exceptional <one-dimensional characters> have $X=\pm1$, $Y=\pm1$.
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