Is there a Lagrangian that produces these equations? How can I find one if it exists?

To answer Question 2: The role of $g$ in Douglas's theorem appears to be to "symmetrize" the potential terms. (I gleaned this from the condition (H1) in the Wikipedia article.) In particular, the fact that $g \Phi = (g \Phi)^T$ implies that multiplying a column vector containing the given equations by $g$ would be fruitful. Doing this yields \begin{align*} \ddot{x}_{A} + \frac{3}{2} \ddot{x}_B + \frac{1}{2}\gamma(x_{A} + x_{B}) &= 0 \\ \frac{3}{2} \ddot{x}_A + 2 \ddot{x}_{B} + \frac{1}{2} \gamma(x_{A} + x_{B}) &= 0 \end{align*} and so it seems that the desired Lagrangian is $$ L = \frac{1}{2} \dot{x}_A^2 + \frac{3}{2} \dot{x}_A \dot{x}_B + \dot{x}_B^2 - \frac{1}{4} \gamma (x_A + x_B)^2. $$

As an aside, note that the above "kinetic term" is not positive definite. This may or may not be relevant for your purposes.

For future reference I recite your differential equations \begin{align} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{A}} - \frac{\partial L}{\partial x_{A}} = \ddot{x}_{A} - \gamma(x_{A} + x_{B}) \tag{EL1}\\ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}_{B}} - \frac{\partial L}{\partial x_{B}} = \ddot{x}_{B} + \gamma(x_{A} + x_{B})\tag{EL2} \end{align}

Unfortunately I am not familiar with Douglas theorem. However I was able to find the following proof which gives an answer to Question 1:

If we assume our Lagrangian to be of the most general form $L=L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)$ and we plug it in the first equation $(EL1)$ collecting the $\ddot{x}_A$ terms yields $$ \frac{\partial^2 L}{\partial \dot{x}_{A}^2}=1 $$ Thus, $L$ has to be a polynomial of second order in $\dot{x}_{A}$, or more specifically $$ L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)=\frac{\dot{x}_A^2}{2}+\dot{x}_A\cdot L_1(\dot{x}_B,x_A,x_B,t)+L_0(\dot{x}_B,x_A,x_B,t) $$ Now we plug this into the second equation $(EL2)$ and collect the $\ddot{x}_A$ and $\ddot{x}_B$ terms and see that $$ \ddot{x}_A:\qquad\frac{\partial L_1}{\partial \dot{x}_{B}}=0\\ \ddot{x}_B\dot{x}_A: \qquad\frac{\partial^2 L_1}{\partial \dot{x}_{B}^2}=0\\ \ddot{x}_B:\qquad\frac{\partial^2 L_0}{\partial \dot{x}_{B}^2}=1 $$ which yields more refined representations of the functions $L0$ and $L1$, namely $$ L(\dot {x}_A,\dot{x}_B,x_A,x_B,t)=\frac{\dot{x}_A^2+\dot{x}_B^2}{2}+\dot{x}_A\cdot \alpha(x_A,x_B,t)+\dot{x}_B\cdot \beta(x_A,x_B,t)+\lambda(x_A,x_B,t) $$ Once again, plugging in the adjusted form in the two equations $(EL1)$ and $(EL2)$ we end up with three equations for the unknown functions $\alpha,\beta$ and $\lambda$. \begin{align} \frac{\partial \alpha}{\partial x_B}&=\frac{\partial \beta}{\partial x_A}\tag{1}\\ \frac{\partial \alpha}{\partial t}-\frac{\partial \lambda}{\partial x_A}&=-\gamma(x_A+x_B)\tag{2}\\ \frac{\partial \beta}{\partial t}-\frac{\partial \lambda}{\partial x_B}&=\gamma(x_A+x_B)\tag{3} \end{align} These equations look promising at first, however if we consider $$ \frac{\partial}{\partial x_B}\big(2)\big)-\frac{\partial}{\partial x_A}\big((3)\big)\\ \Leftrightarrow \frac{\partial^2\alpha}{\partial t\partial x_B}-\frac{\partial^2\beta}{\partial t\partial x_A}=-2\gamma\\ \stackrel{(1)}{\Rightarrow}0=-2\gamma $$ we see that a solution can only exist for the trivial case of $\gamma=0$.

edit: for the last step I considered $\alpha,\beta$ and $\gamma$ to be differentiable twice. If we relax this condition we might find a solution to the system $(1)-(3)$.