= Solution
The exponential input has norm squared $\exp(|\alpha_1|^2+|\alpha_2|^2)$. Its normalized form is a product of <coherent states>. Substitution of the inverse transformation gives
$$
\alpha_1a_1^\dagger+\alpha_2a_2^\dagger
=\beta_1b_1^\dagger+\beta_2b_2^\dagger,\qquad
\beta_1=\frac{\alpha_1+\alpha_2}{\sqrt2},\quad
\beta_2=\frac{\alpha_1-\alpha_2}{\sqrt2}.
$$
Since the output <creation operators> commute, the normalized output factorizes as $|\beta_1\rangle|\beta_2\rangle$. For all nonnegative integers $n_1,n_2$,
$$
\boxed{P(n_1,n_2)=
e^{-|\beta_1|^2-|\beta_2|^2}
\frac{|\beta_1|^{2n_1}}{n_1!}
\frac{|\beta_2|^{2n_2}}{n_2!}.}
$$
These are independent <Poisson distributions>, not a fixed-total-particle-number <distribution>. <unitarity> gives $|\beta_1|^2+|\beta_2|^2=|\alpha_1|^2+|\alpha_2|^2$. Equal complex inputs make $\beta_2=0$, so that port is certainly empty; opposite inputs empty port 1. This is amplitude <interference> of coherent inputs, distinct from the fixed-number two-particle <interference> in (c).
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