Roots in Thompson's groups Let $G \in \{F,T,V\}$, and let $g$ be an element of Thompson's $G$.

Question 1. When does $g$ have an $n$th root?

If this is a complicated issue, and a description is not so easy to give, I am especially interested in the following more specific questions (for any of the $G$).

Question 2. Given $(g,n)$, is it decidable whether $g$ has an $n$th root?


Question 3. Can an element $g$ have roots of arbitrarily large orders?

For $G = F$, if $g \neq 1_F$, looking at the slope at the leftmost point of $g$'s support we get an upper bound on the possible $n$'s, so Question 3 has a negative answer for this one. For $T$ and $V$ I don't see any trivial argument. I feel like Question 2 should have an easy "yes" answer, but I don't know how to prove that.
 A: About Question 3, the answer is "no" because Thompson's groups $F$, $T$, and $V$ act properly on CAT(0) cube complexes (as proved by D. Farley, see MR1978047 and MR2136028).
Indeed, a theorem due to F. Haglund (see arxiv:0705.3386) shows that every isometry $g \in \mathrm{Isom}(X)$ of a CAT(0) cube complex $X$ with unbounded orbits acts as a translation on some bi-infinite combinatorial geodesic, and that $\|g^n\|=|n| \cdot \|g\|$ for every $n \in \mathbb{Z}$ where $\|\cdot\|$ denotes the combinatorial translation length of the isometry under consideration. Therefore, if $G$ acts properly on $X$ and if $g \in G$ is an infinite order element, then $g$ cannot be an $n$th root if $n$ is larger than $\|g\|$.
A: For $F$, all questions are answered by Theorem 7.1 of this: https://arxiv.org/pdf/0709.1987.pdf
For questions 1 and 2 it just amounts to checking whether numbers divide other numbers, and for question 3 the answer is "no".
A: It is possible to use closed strand diagrams to check whether a given element of $F$, $T$, or $V$ has an $n$th root.  Before discussing this algorithm, I'd like to say a little bit about the situation in general.  The main result for Thompson's group $F$ is the following.
Theorem 1. Let $f\in F$.  Then there exists a unique $g\in F$ so that $g$ is a root of $f$ and every root of $f$ is a power of $g$.
It follows easily an element of $F$ can have an $n$th root for only finitely many different $n$, and that for each such $n$ the root is unique.  Matthew Brin likes to refer to the element $g$ above as the "rootiest root" for the element $f$.  I believe the above result appears in the article "Presentations, conjugacy, roots, and centralizers in groups of piecewise linear homeomorphisms of the real line" by Matthew Brin and Craig Squier, and was also proven for arbitrary diagrams groups by Victor Guba and Mark Sapir in "Diagram groups".
The situation in Thompson's groups $T$ and $V$ is quite different.  First we have the following:
Theorem 2. Every element of $T$ or $V$ of finite order is contained in a subgroup isomorphic to $\mathbb{Q}/\mathbb{Z}$.
This theorem was proven for $T$ by Collin Bleak, Martin Kassabov, and Francesco Matucci in "Structure theorems for groups of homeomorphisms of the circle" (published version | arXiv version), and the proof for $V$ is similar.  It follows from the proof that every element of $T$ or $V$ of finite order in fact has infinitely many different $n$th roots for each $n\geq 2$, and is hence contained in uncountably many different subgroups isomorphic to $\mathbb{Q}/\mathbb{Z}$.
On the other hand, infinite-order elements of $T$ and $V$ have very few roots:
Theroem 3. Let $f$ be an element of $T$ or $V$ of infinite order.  Then $f$ has an $n$th root for only finitely many $n\in\mathbb{N}$.
Anthony Genevois gives a proof of this statement in his answer using CAT(0) complexes, and it also follows from the strand diagrams approach.
Uniqueness of roots varies from element to element.  For example, we have:
Theorem 4. Let $f$ be an element of $F$ whose only fixed points are $0$ and $1$.  Then the roots of $f$ in $T$ or $V$ are the same as the roots of $f$ in $F$.
Proof: Suppose without loss of generality that $0$ is a replling fixed point and $1$ an attracting fixed point for $f$, and let $g$ be any root of $f$ in $V$.  Then $1$ is also an attracting fixed point for $g$, with basin $(0,1]$.  Clearly $g$ is order-preserving in some neighborhood $U$ of $1$.  Since $f$ and $g$ commute and $f$ is order-preserving, it follows that $g$ is order preserving on $f^{-n}(U)$ for all $n\in\mathbb{N}$.  But $\bigcup_{n\in\mathbb{N}}f^{-n}(U) = (0,1]$, so $g$ is order-preserving on $(0,1]$ and hence lies in $F$.$\quad\square$
On the other hand, we have:
Theorem 5. There exist elements of $F$ that have infinitely many different square roots in $T$.
Proof: For $f,g\in F$, let $f\oplus g$ denote the element which acts as $f$ on $[0,1/2]$ and $g$ on $[1/2,1]$, i.e.
$$
(f\oplus g)(x) = \begin{cases}\tfrac{1}{2}f(2x) & \text{if }x\in [0,1/2] \\[3pt] \tfrac{1}{2}+\tfrac{1}{2}g(2x-1) & \text{if }x\in (1/2,1].\end{cases}
$$
If $f$ is any nontrivial element of $F$, I claim that $f\oplus f$ has infinitely many square roots in $T$.  To see this, let $\delta\in T$ be the half-rotation of the circle, which switches the intervals $[0,1/2]$ and $[1/2,1]$.  Then $(f^k\oplus f^{1-k})\delta$ is a square root of $f\oplus f$ for any $k\in\mathbb{Z}$, since
$$
(f^k\oplus f^{1-k})\delta(f^k\oplus f^{1-k})\delta = (f^k\oplus f^{1-k})(f^{1-k}\oplus f^k) = f\oplus f.\tag*{$\square$}
$$
Note that the elements $(f^k\oplus f^{1-k})\delta$ in the above proof are all conjugate by powers of $1\oplus f$.  In general, the $n$th roots of an element of $T$ or $V$ ought to fall into finitely many different conjugacy classes.
Roots in $F$ Using Strand Diagrams
Closed strand diagrams were introduced by myself and Francesco Matucci in the paper "Conjugacy and dynamics in Thompson's groups" (published version | arXiv version) to solve the conjugacy problems in $F$, $T$, and $V$.  Conjugacy is closely related to roots since having an $n$th root is a conjugacy invariant, and it is possible to use closed strand diagrams to check whether an element of $F$, $T$, and $V$ have an $n$th root.
As described in my paper with Francesco Matucci, the conjugacy classes in $F$ are in one-to-one correspondence with certain geometric objects calles "reduced annular strand diagrams".

This is a certain kind of directed graph embedded in the annulus up to isotopy.  There is a simple algorithm to compute the reduced annular strand diagram for an element of $F$ involving first gluing together the two trees of a tree pair diagram and then iteratively performing certain reduction moves.
The connection between reduced annular strand diagrams and roots is the following.
Observation: If $f\in F$ and $n\geq 1$, the reduced annular strand diagram for $f^n$ is the $n$-fold cover of the reduced annular strand diagram for $f$.
For example, here are the reduced annular strand diagrams for a certain element $f\in F$ and its cube $f^3$:
         
Corollary. An element of $F$ has an $n$th root if and only if its reduced annular strand diagram can be drawn to have $n$-fold rotational symmetry.
It is easy to check for such symmetry algorithmically.  For example, for each component we can enumerate its graph automorphisms and check whether there exists an automorphism of order $n$ that preserves the counterclockwise order of edges at each vertex such that the orbit of each vertex has size $n$.
By the way, a version of this "rotational symmetry" algorithm to check whether an element of $F$ has an $n$th root was first described by Brin and Squier, and was also described by Guba and Sapir for arbitrary diagram groups.
Roots in $T$ and $V$
All of this ought to work just fine for $T$ and $V$, though as far as I know the details haven't been written up anywhere. My paper with Francesco Matucci describes closed strand diagrams for $T$ and $V$ which are analogs of the reduced annular strand diagrams for $F$.  We refer to these as "reduced toral strand diagrams" for $T$ and "reduced abstract closed strand diagrams" for $V$.  Checking whether an element of $T$ or $V$ has an $n$th root is equivalent to checking whether the reduced closed diagram is an $n$-fold cover of a smaller diagram.
There is a slight complication involving free loops, since the $n$-fold cover of a closed diagram that involves free loops might not be reduced.  (Specifically, it might have duplicate copies of the free loops that need to be combined.)  For $T$ this is only a problem for finite-order elements, which is ok since we already know that such elements have $n$th roots.  For $V$ the right thing to do is to simply discard all of the free loops in a reduced abstract closed strand diagram before checking for $n$-fold symmetry.
A: More or less to add to Jim's answer: The strand diagram material is equivalent to Brin's Revealing Pair technology (introduced in his paper Brin's Higher Dimensional Thompson groups where he introduced $2V$).  We use this revealing pair technology in fashion very similar to Jim's description for $F$ elements above to describe roots of elements in $V$ in (mostly Section 7) of our paper "Centralizers in the R. Thompson group $V_n$" published in GGD in 2013 (see Centralizer's in $V$).
The main point is that the centralizer of a non-torsion element is virtually $Z$ over the region of the support of the element in Cantor space.  So, there is a finite set of "rootiest roots" supported there.  If our element $g$ has root $r$ in this set, and $r^k=g$, then away from the support of $r$, one could put an infinitude of torsion elements of order dividing $k$.  All of these created elements are also "roots" of $g$, but in a stupid way.
Ewa Bieniecka in her thesis developed and generalized the revealing pair technology further, and answered the other direction of the results characterized in our centralizer's paper.  I think this is a very gentle resource for learning flow graphs and revealing pairs (the crucial technologies).
