= Solution
Let $x_<=\min(x,x')$ and $x_>=\max(x,x')$. A homogeneous solution vanishing at the left wall is $u(x)=\sin[k(x+a/2)]$; one vanishing at the right wall is $v(x)=\sin[k(a/2-x)]$. Their <Wronskian> is $uv'-u'v=-k\sin ka$. Continuity and the source <derivative> jump $G'_+-G'_-=-2m/\hbar^2$ therefore give
$$
\boxed{G(x,x')=\frac{2m}{k\hbar^2\sin ka}
\sin[k(x_<+a/2)]\sin[k(a/2-x_>)].}
$$
This is exactly the printed form because $x_<= (x+x'-|x-x'|)/2$ and $x_>= (x+x'+|x-x'|)/2$, with a minus sign from reversing the argument of the second <sine>.
The true poles occur at $k_n=n\pi/a$, $n=1,2,\ldots$, corresponding to
$$
\boxed{E_n=\frac{\hbar^2\pi^2n^2}{2ma^2}.}
$$
The apparent zero at $k=0$ is removable: the numerator and $k\sin ka$ both vanish quadratically. Negative values of $k_n$ duplicate the same energies, rather than describing extra levels. The <boundary conditions> and source jump uniquely determine the inverse whenever $E$ is not an <eigenvalue>.
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