= Solution
Assume $a>0$ and put $\mu=mg/\hbar^2$, $q=k/\hbar$, and $\gamma=\mu/q$. At a <delta potential> located at $x_0$, integration of the <Schrödinger equation> gives the matching conditions
$$
\psi(x_0^+)=\psi(x_0^-),\qquad \psi'(x_0^+)-\psi'(x_0^-)=2\mu\psi(x_0).
$$
Writing the adjacent <plane wave> coefficients as $(A_L,B_L)$ and $(A_R,B_R)$ gives
$$
(A_R-A_L)e^{iqx_0}+(B_R-B_L)e^{-iqx_0}=0,
$$
$$
iq\bigl[(A_R-A_L)e^{iqx_0}-(B_R-B_L)e^{-iqx_0}\bigr]=2\mu(A_Le^{iqx_0}+B_Le^{-iqx_0}).
$$
Solving these two linear equations yields the <transfer matrix of a delta potential>:
$$
\begin{pmatrix}A_R\\B_R\end{pmatrix}=D(x_0)\begin{pmatrix}A_L\\B_L\end{pmatrix},\qquad D(x_0)=\begin{pmatrix}1-i\gamma&-i\gamma e^{-2iqx_0}\\i\gamma e^{2iqx_0}&1+i\gamma\end{pmatrix},\qquad\det D(x_0)=1.
$$
For the <symmetric double-delta potential>, use separate <plane wave> pairs in $x<-a$, $-a<x<a$, and $x>a$. Successive matching gives $M=D(a)D(-a)$ between the two exterior pairs. For real $q$, multiplication gives
$$
M=\begin{pmatrix}d^*&-iF\\iF&d\end{pmatrix},\qquad d=(1+i\gamma)^2+\gamma^2e^{4iqa},\qquad F=2\gamma\bigl[\cos(2qa)+\gamma\sin(2qa)\bigr].
$$
The <determinant> identity $\det M=1$ reads $|d|^2-F^2=1$. Left incidence means $(t,0)^T=M(1,r)^T$, whence
$$
\boxed{t=\frac1d=\frac{q^2}{(q+i\mu)^2+\mu^2e^{4iqa}},\qquad r=-\frac{iF}{d}.}
$$
Here the first relation also follows from $t=M_{11}-M_{12}M_{21}/M_{22}=\det M/M_{22}$. Therefore the requested <transmission probability> and <quantum reflection probability>, as functions of the printed <momentum> $k$, are
$$
\boxed{T(k)=\frac{1}{1+F(k)^2},\qquad R(k)=\frac{F(k)^2}{1+F(k)^2},\quad F(k)=\frac{2\mu\hbar}{k}\left[\cos\frac{2ka}{\hbar}+\frac{\mu\hbar}{k}\sin\frac{2ka}{\hbar}\right].}
$$
They obey $R+T=1$. As checks, $g=0$ gives $T=1$, and the coincident limit $a\to0$ gives $T=q^2/(q^2+4\mu^2)$, the answer for a single <delta potential> of strength $2g$.
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