Over ${\bf C}$, An easy counterexample to question 3 is $f(x) = x^2$, $g(x) = cx^2$ where $c$ is a nontrivial cube root of unity. Then $f(f(x)) = g(g(x)) = x^4$ but $f$ and $g$ do not commute. There are similar examples for higher iterates.
[Added later] A more exotic construction yields further examples, some defined over ${\bf Q}$, such as the degree-4 pair $$ f(y) = \frac{y^4+18y^2-47}{8y^3}, \phantom{\infty} g(y) = \frac{f-3}{f+1} = \frac{y^4-24y^3+18y^2-27}{y^4+8y^3+18y^2-27} $$ with $f \circ f = g \circ g$ but $f \circ g \neq g \circ f$. This is a "Lattès map" associated to the elliptic curve $E: y^2 = x^3 + 1$: the function $f$ comes from the doubling map $P \mapsto 2P$, and $g$ comes from $P \mapsto 2P+T$ where $T$ is the 3-torsion point $(0,1)$ (as the $(f,g)=(x^2,cx^2)$ example does on the multiplicative group). This elliptic curve yields examples of $f \circ f = g \circ g$ and $f \circ g \neq g \circ f$ with any degree $m^2+mn+n^2$ as long as that's not a multiple of 3, with $f,g \in {\bf Q}(y)$ if $n=0$. Other elliptic curves with complex multiplication yield further examples using the $x$-coordinate rather than the $y$-coordinate, e.g. $f(x) = -x(x^4+6x^2-3)^2 / (3x^4-6x^2-1)^2$ and $g = (f-1)/(f+1)$ from tripling on $y^2=x^3-x$.
