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I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens, with probability density $dP_A/d\tau$. Then from $\tau$ to $T$ the event $B$ does not happen, with probability $e^{-(T-\tau)\lambda}$. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T \frac{dP_A(\tau)}{d\tau}e^{-(T-\tau)\lambda}\,d\tau.$$

I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens, with probability density $dP_A/d\tau$. Then from $\tau$ to $T$ the event $B$ does not happen, with probability $e^{-(T-\tau)\lambda}$. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T \frac{dP_A(\tau)}{d\tau}e^{-(T-\tau)\lambda}\,d\tau$$ $$\Rightarrow P_{A\rightarrow B}=\lambda\int_0^T e^{-(T-\tau)\lambda}P_A(\tau) \,d\tau.$$

I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens. We require a time interval from 0 to $\tau$ where $A$ does not happen [probability $1-P_A(\tau)$], then $A$ happens [probability density $p_A(\tau)$], and then from $\tau$ to $T$ the event $B$ does not happen [probability $e^{-(T-\tau)\lambda}$]. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T [1-P_A(\tau)]p_A(\tau)e^{-(T-\tau)\lambda}\,d\tau.$$

The extra information you need, the probability $p_A(t)dt$ that event $A$ happens in the time interval $(t,t+dt)$, is required -- it does not follow from $P_A(t)$ in the general case. If $A$ is also a Poisson process, with intensity $\lambda'$, then $P_A(t)=1-e^{-\lambda' t}$ and $p_A(t)=\lambda'$.

I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens, with probability density $dP_A/d\tau$. Then from $\tau$ to $T$ the event $B$ does not happen, with probability $e^{-(T-\tau)\lambda}$. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T \frac{dP_A(\tau)}{d\tau}e^{-(T-\tau)\lambda}\,d\tau.$$

I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens. We require a time interval from 0 to $\tau$ where $A$ does not happen [probability $1-P_A(\tau)$], then $A$ happens [probability $dP_A/d\tau$], and then from $\tau$ to $T$ the event $B$ does not happen [probability $e^{-(T-\tau)\lambda}$]. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T [1-P_A(\tau)]\frac{dP_A(\tau)}{d\tau}e^{-(T-\tau)\lambda}\,d\tau.$$

I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens. We require a time interval from 0 to $\tau$ where $A$ does not happen [probability $1-P_A(\tau)$], then $A$ happens [probability density $p_A(\tau)$], and then from $\tau$ to $T$ the event $B$ does not happen [probability $e^{-(T-\tau)\lambda}$]. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T [1-P_A(\tau)]p_A(\tau)e^{-(T-\tau)\lambda}\,d\tau.$$

The extra information you need, the probability $p_A(t)dt$ that event $A$ happens in the time interval $(t,t+dt)$, is required -- it does not follow from $P_A(t)$ in the general case. If $A$ is also a Poisson process, with intensity $\lambda'$, then $P_A(t)=1-e^{-\lambda' t}$ and $p_A(t)=\lambda'$.