= Solution
Put
$$
\kappa=\frac{\sqrt{2m(V_0-E)}}{\hbar}.
$$
An even solution is proportional to $\cosh(\kappa x)$ inside the <potential barrier>. Normalize the right exterior region as
$$
\psi(x)=e^{-ikx}+(r+t)e^{ikx}\qquad(x>a/2).
$$
Continuity of $\psi$ and $\psi'$ at $x=a/2$ gives, with
$\gamma=\kappa\tanh(\kappa a/2)$,
$$
\frac{ik\bigl((r+t)e^{ika/2}-e^{-ika/2}\bigr)}
{(r+t)e^{ika/2}+e^{-ika/2}}=\gamma.
$$
Solving,
$$
\boxed{r+t=e^{-ika}\frac{ik+\gamma}{ik-\gamma}},
\qquad
\gamma=\kappa\tanh(\kappa a/2).
$$
Its modulus is one, as expected for the even-parity eigenvalue of the unitary S-matrix.
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