TITLE: solving $f(f(x))=g(x)$
QUESTION [102 upvotes]: This question is of course inspired by the question How to solve f(f(x))=cosx
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a bounded interval. [EDIT: actually he can do rather better than this, solving the equation away from a bounded interval (with positive measure)].
I've always found such questions ("solve $f(f(x))=g(x)$") rather vague because I always suspect that solutions are highly non-unique, but here are two precise questions which presumably are both very well-known:
Q1) Say $g:\mathbf{R}\to\mathbf{R}$ is an arbitrary function. Is there always a function $f:\mathbf{R}\to\mathbf{R}$ such that $f(f(x))=g(x)$ for all $x\in\mathbf{R}$?
Q2) If $g$ is as above but also assumed continuous, is there always a continuous $f$ as above?
The reason I'm asking is that these questions are surely standard, and perhaps even easy, but I feel like I know essentially nothing about them. Apologies in advance if there is a well-known counterexample to everything. Of course Q1 has nothing to do with the real numbers; there is a version of Q1 for every cardinal and it's really a question in combinatorics.
EDIT: Sergei Ivanov has answered both of these questions, and Gabriel Benamy has raised another, which I shall append to this one because I only asked it under an hour ago:
Q3) if $g$ is now a continuous function $\mathbf{C}\to\mathbf{C}$, is there always continuous $f$ with $f(f(x))=g(x)$ for all $x\in\mathbf{C}$?
EDIT: in the comments under his answer Sergei does this one too, and even gives an example of a continuous $g$ for which no $f$, continuous or not, can exist.
Related MO questions: f(f(x))=exp(x) and other functions just in the middle between linear and exponential, and Does the exponential function has a square root.

REPLY [6 votes]: The following is inspired by the chapter Partition Polynomials in Riordan's Combinatorial Identities where $g(x)$ is analytic and $g(0)=0$. Find the Taylor series of $f(x)$ at zero by evaluating the derivatives of $f(f(x))=g(x)$ at $0$ in succession.
Set $f(0)=0$, giving $f(f(0))=g(0)=0$.
The first derivative gives $f'(x) f'(f(x))=g'(x)$, therefore $f'(0)=\sqrt{g'(0)}$ and $f'(0)=-\sqrt{g'(0)}$. Set $f'(0)=\sqrt{g'(0)}$ for this example. 
The second derivative $f'(x)^2 f''(f(x))+f'(f(x)) f''(x)=g''(x)$ produces $f''(0)=\frac{g''(0)}{g'(0)+\sqrt{g'(0)}}$.
The first few term of the Taylor series are
$f(x)=\sqrt{g'(0)}x+\frac{ g''(0)}{2
   \left(g'(0)+\sqrt{g'(0)}\right)}x^2$ $+\frac{
   \left(-3 g''(0)^2+g^{(3)}(0) g'(0)^{3/2}+2
   g^{(3)}(0) g'(0)+g^{(3)}(0) \sqrt{g'(0)}\right)}{6
   \left(\sqrt{g'(0)}+1\right)^2 g'(0)
   \left(g'(0)+1\right)}x^3+O(4)$.
This solution is based on $g(x)$ having a fixed point. Technically the question stated is whether there is always a solution and is focused on counter examples instead of a broad general answer as I have provided. But it does raise the question of whether there is a connection between $g(x)$ not having a fixed point and the counter examples that the other authors have provided.