= Solution
Use the normalized <chordal metric>
$$
\boxed{\chi(z,w)=\frac{|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}},\qquad\chi(z,\infty)=\frac1{\sqrt{1+|z|^2}},\qquad\chi(\infty,\infty)=0.}
$$
Under <stereographic projection> it is half the Euclidean chord distance on the unit sphere, which proves the metric axioms and the agreement with the usual sphere topology. If $\rho$ is spherical <geodesic> distance, then $\rho=2\arcsin\chi$, so $2\chi\leq\rho\leq\pi\chi$.
A <rational map> is holomorphic, hence smooth, as a map of <compact> spheres. At a <pole> use $1/R$ as target coordinate, and at infinity use $1/z$ as source coordinate; neither is a singularity of the sphere map. Its differential in the round metric has a finite maximum norm $L$. Integrating its norm along a minimizing spherical <geodesic> gives $\rho(Rz,Rw)\leq L\rho(z,w)$. The preceding comparisons imply
$$
\boxed{\chi(Rz,Rw)\leq\frac{\pi L}{2}\chi(z,w).}
$$
This proves that the <rational map is Lipschitz in the chordal metric>. For a constant map take constant zero; for a nonconstant map the bound is finite by <compactness>. In finite coordinates the differential norm is $R^\#(z)=|R'(z)|(1+|z|^2)/(1+|R(z)|^2)$, with its continuous sphere extension.
Now suppose $R,S$ are <commuting maps>. We first prove $S(F(R))\subseteq F(R)$. The family $\{R^n\}$ is locally <equicontinuous> on $F(R)$, by the normality equivalence in question 1. Commutation and the Lipschitz estimate give, for all $n$,
$$
\chi(R^n(S(x)),R^n(S(x_0)))=\chi(S(R^n(x)),S(R^n(x_0)))\leq C_S\chi(R^n(x),R^n(x_0)).
$$
For $x_0\in F(R)$ choose a source ball within that <open set>, small enough to make the right side less than any prescribed $\epsilon$ uniformly in $n$. Holomorphic openness makes its image contain a ball about $S(x_0)$. Every point $w$ of that target ball has some preimage $x$ in the chosen source ball, so the same inequality proves <equicontinuity> of the iterates of $R$ at $S(x_0)$. Repeating at each point of the image gives local normality on that image. Thus the stated inclusion holds even where $S$ has a <rational critical point>.
Carefully state the permitted <Montel theorem>: a family of meromorphic maps on an open domain, all omitting the same three distinct values of the <Riemann sphere>, is normal for locally uniform spherical convergence. Since $\deg R\geq2$, question 1 shows that $J(R)$ contains at least three distinct points. The forward inclusion just proved means every iterate of $S$ maps $F(R)$ into itself and therefore omits those three points there. Montel gives $F(R)\subseteq F(S)$. Exchanging the roles, using $\deg S\geq2$, gives the opposite inclusion. Consequently
$$
\boxed{F(R)=F(S),\qquad J(R)=J(S).}
$$
This proves that <commuting rational maps of degree at least two have the same Julia set> without assuming either is an iterate of the other.
For <commuting maps> of degree one with different <Julia sets>, take
$$
\boxed{R(z)=2z,\quad S(z)=z/2;\qquad J(R)=\{0\},\quad J(S)=\{\infty\}.}
$$
Their two compositions are the identity, and the <Julia sets> follow from the expanding and contracting Möbius cases in question 1.
For equal <Julia sets> without commutation, take
$$
\boxed{R(z)=z/2,\quad S(z)=z/2+1;\qquad J(R)=J(S)=\{\infty\}.}
$$
Both affine maps contract toward their finite <fixed points> and have a <repelling periodic point> at infinity. But $R\circ S=z/4+1/2$, while $S\circ R=z/4+1$, so they do not commute.
Finally the distinct <Chebyshev polynomials>
$$
\boxed{R(z)=2z^2-1,\qquad S(z)=4z^3-3z}
$$
have degrees two and three and neither is a pure power of $z$. Direct expansion gives both compositions equal to $32z^6-48z^4+18z^2-1$. Equivalently $T_m(T_n(z))=T_{mn}(z)$, first seen at $z=\cos\theta$ and then as a <polynomial identity>. This provides the requested higher-degree pair.
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