For any primitive element $a$ in a free group of rank two we have $a^k ba^l=b$ only if $(k,l)=(0,0)$ provided $b\not\in \langle a\rangle$ 
Problem 1: Let $F$ be a free group of rank at least two, and $a, b\in F$ be two
non-trivial elements with $b\not \in \langle a\rangle$. Suppose,  $a\neq x^n$ for any $x\in F$ and any
$n\geq 2$, i.e. $a$ is primitive. Then, $a^kb a^l=b\iff (k,l)=(0,0)$.

Note that $a$ is primitive is necessary, otherwise, $a$ could be $b^m$ for some $m\in \Bbb Z$, and then $a^kba^l=b^{mk}bb^{ml}=b^{m(k+l)+1}$ would equal to $b$ for each $k=-l$.
Let $F$ be free on a set $\{g_i:i\in \Lambda\}$ of generators. Now, write $a$ and $b$ as reduced words. If there exist two distnict elements $g_a, g_b\in \{g_i:i\in \Lambda\}$ having contributions in the reduced representations of $a,b$, respectively, then $a^k ba^l\neq b$ unless $(k,l)=(0,0)$, and we do not need the assumption $a$ is primitive in this case.
So, we need to consider the case when both $a$ and $b$ have reduced representations obtained from the same set.

Problem 2: Suppose I replace $F$ by this non-free group $\big\langle x_1,x_2,y_1,y_2\big|[x_1,y_1]=[x_2,y_2]\big\rangle$
keeping fixed all other hypothesis on $a,b$. Is it again true that
$a^k ba^l=b$ only if $(k,l)=(0,0)$?

 A: Here is a quick proof, assuming that you know how to solve the conjugacy problem for free groups:
Note that if elements $x, y\in F$ commute then $\langle x, y\rangle$ is cyclic, which can be proven using the solution to the conjugacy problem for $F$ (or by applying the nuclear bomb which says that "subgroups of free groups are free").
Now, assume that $a^kba^l=b$ with $k,l\neq0$, where $a$ and $b$ are as in the question. The identity $a^kba^l=b$ implies that $a^k$ and $a^{-l}$ are conjugate. By applying the solution to the conjugacy problem, we see that $k={-l}$, and hence $a^k$ and $b$ commute. As subgroups of free groups are free, we have that $\langle a^k, b\rangle$ is cyclic, and in fact $\langle a\rangle=\langle a^k, b\rangle$. Hence, $b\in\langle a\rangle$, a contradiction.

Problem 2 also holds true. There are a number of ways of seeing this, all depending on what you know. I cannot see an easy, no-prior-knowledge proof though. The "slickest" way is perhaps to note that this is a surface group, and subgroups of surface groups are either surface groups of higher genus or free, and so $\langle a, b\rangle$ is free. Another way is to note that this result is true for all torsion-free hyperbolic groups (the standard phrase is "maximal cyclic subgroups are malnormal"), and the group here is a torsion-free hyperbolic group.
