= Solution
Specify the amplitude convention. For incident stress/pressure amplitude one, the normal-incidence <seismic reflection coefficient> and stress <seismic transmission coefficient> are
$$
R=\frac{Z_b-Z_a}{Z_b+Z_a},\qquad T_\sigma=1+R=\frac{2Z_b}{Z_a+Z_b}.
$$
At limestone–sandstone, \b[$R_P=-3/19=-0.1579$ and $T_{\sigma P}=16/19=0.8421$]. At sandstone–salt, \b[$R_P=0$ and $T_{\sigma P}=1$]. The first return reverses the chosen stress-pulse polarity; the second interface produces no normal-incidence P return.
For a particle-velocity amplitude convention, $R_v=-R$ and $T_v=1-R$, giving $(R_v,T_v)=(0.1579,1.1579)$ and $(0,1)$ respectively. These are amplitude coefficients, not fractions of <energy>. The lossless reflected and transmitted energy fractions are $\mathcal R=R^2$ and $\mathcal T=4Z_aZ_b/(Z_a+Z_b)^2=1-R^2$, which sum to one.
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