= Solution
In the interaction picture,
$$
a_f^{(1)}(t)=-\frac{i\lambda}{\hbar}\sum_{k=1}^2\alpha_kW_k\int_0^te^{(i\omega_{fi_k}-\gamma)t'}\cos(\omega t')dt'.
$$
At infinity this is $-(i\lambda/\hbar)\sum_k\alpha_kW_kF_k$, where
$$
F_k=\tfrac12\left[\frac1{\gamma-i(\omega_{fi_k}+\omega)}+\frac1{\gamma-i(\omega_{fi_k}-\omega)}\right],
$$
so $b_1=1,b_2=1,b_3=-1$ and $\omega_{fi_k}=(E_f-E_{i_k})/\hbar$. The probability is $(\lambda^2/\hbar^2)|\sum_k\alpha_kW_kF_k|^2$, displaying interference between initial components. For $N$ equal response factors, Cauchy-Schwarz gives $|\sum_k\alpha_k|^2\le N$, attained by equal phases and $|\alpha_k|=N^{-1/2}$; the largest probability is $N\lambda^2|WF|^2/\hbar^2$ within perturbation theory.
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