= Solution
Fix a separable closure $k^s$. The <Tate module> and <rational Tate module> are
$$
T_\ell(E)=\varprojlim_n E[\ell^n](k^s),\qquad V_\ell(E)=T_\ell(E)\otimes_{\mathbb Z_\ell}\mathbb Q_\ell,
$$
where the transition maps are multiplication by $\ell$. Since $\ell$ is different from the characteristic, $E[\ell^n](k^s)\simeq(\mathbb Z/\ell^n)^2$ and these transition maps are surjective. Choosing compatible bases gives $T_\ell(E)\simeq\mathbb Z_\ell^2$, and $T_\ell(E)/\ell^nT_\ell(E)\simeq E[\ell^n]$. Both spaces carry the natural continuous action of the <absolute Galois group> of $k$ and of the base-defined <endomorphism ring of an elliptic curve>.
For \b[(i)], choose a nonrational $a\in R$. Such an element exists because the intersection of the given two-dimensional rational subalgebra with the integral <endomorphism ring> contains a full lattice. The characteristic equation of an elliptic <endomorphism> is
$$
a^2-\operatorname{tr}(a)a+\deg(a)=0.
$$
Since $a\in K$ is nonrational, its minimal polynomial over $\mathbb Q$ has degree two. Comparing the two monic equations gives $\operatorname{tr}(a)=a+\bar a$ and $\deg(a)=a\bar a$. The same is the characteristic polynomial on the <rational Tate module>: the <Weil pairing> gives determinant $\deg(a)$, and applying it to $1-a$ gives trace $1+\deg(a)-\deg(1-a)$. Thus on $V_\ell(E)$ the characteristic polynomial is
$$
(X-a)(X-\bar a).
$$
If $K_\ell$ is a quadratic <field> over $\mathbb Q_\ell$, a unital $K_\ell$-action on this two-dimensional $\mathbb Q_\ell$-space makes it a one-dimensional $K_\ell$-space. If $K_\ell\simeq\mathbb Q_\ell\times\mathbb Q_\ell$, the two roots above are distinct in $\mathbb Q_\ell$ and their eigenspaces each have dimension one. The two idempotent factors therefore have equal rank one, so the <module> is again free of rank one. This also proves faithfulness in the split case. Hence
$$
\boxed{V_\ell(E)\simeq K_\ell\quad\text{as }K_\ell\text{-modules}.}
$$
For \b[(ii)], first prove an integral divisibility statement. If $a\in R$ maps $T_\ell(E)$ into $\ell^nT_\ell(E)$, then its induced action on $E[\ell^n]$ is zero. The <kernel of an isogeny> of $[\ell^n]$ is therefore contained in the kernel of $a$. The quotient property of this <isogeny of elliptic curves> gives a unique <endomorphism> $b$ with
$$
a=b\circ[\ell^n]=[\ell^n]\circ b.
$$
This factorization is defined over $k$: applying any descent automorphism preserves it, and uniqueness fixes $b$. Inside the rational <endomorphism ring>, $b=a/\ell^n$ belongs to $K$, so $b\in R$. Consequently
$$
\{a\in R:aT_\ell(E)\subseteq\ell^nT_\ell(E)\}=\ell^nR.
$$
Here the prime-to-characteristic torsion is étale, so vanishing on its geometric points is vanishing on the entire kernel scheme.
Every element of $R_\ell$ preserves the <Tate module> by continuity. Conversely suppose $\phi\in K_\ell$ preserves it. Choose $n\geq0$ with $\ell^n\phi\in R_\ell$, and approximate it by $a\in R$ modulo $\ell^nR_\ell$. Write $\ell^n\phi=a+\ell^nc$ with $c\in R_\ell$. Then $aT_\ell(E)\subseteq\ell^nT_\ell(E)$, so the divisibility statement gives $a/\ell^n\in R$. It follows that $\phi=a/\ell^n+c\in R_\ell$. Thus the <CM Tate module stabilizer> is
$$
\boxed{\{\phi\in K_\ell:\phi T_\ell(E)\subseteq T_\ell(E)\}=R_\ell.}
$$
This does not presume freeness of the integral <Tate module> over a nonmaximal order.
For \b[(iii)], every element of $R$ is defined over $k$, so the <Galois representation> commutes with $R$ and therefore with $K_\ell$. By (i), its action on $V_\ell(E)$ is multiplication by an element of $K_\ell^\times$. The action and its inverse preserve $T_\ell(E)$; by (ii), this scalar and its inverse belong to $R_\ell$. Hence
$$
\boxed{\rho_\ell(G_k)\subseteq R_\ell^\times,\quad\text{so the image is abelian}.}
$$
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