Let $X,Y$ two independent Poisson processes with parameters $\lambda, \mu$ and $T=\inf\{t\geq 0 \mid Y(t)=1\}$. Calculate $P(X(T/2)=k)$ with $k=0,1, \ldots$.
What I've tried is
By law of total probability \begin{align} P\left(X \Bigl(\frac{T}{2}\Bigr)=k\right)&=\int_0^\infty P\bigl(X(t/2)=k \mid T=t \bigr)P(T=t)\,\mathrm{d}t\\ &=\int_0^\infty \frac{\lambda^k t^k e^{-\lambda t/2}}{2^{k}k!}\mu e^{-\mu t} \,\mathrm{d}t\\ &=\frac{\lambda^k}{2^{k}(\lambda/2+\mu)^{k+1}} \end{align}
but I don't know if I'm using well the total probability law and the fact that the waiting times has Beta distribution (in this case the waiting time is the first, then is exponentially distributed).
Thanks for the advices.
Your work is mostly correct, but you are missing a factor of $\mu$; i.e., $$\begin{align}\Pr[X(T/2) = k] &= \frac{\mu \lambda^k}{2^k (\lambda/2 + \mu)^{k+1}} \\ &= \frac{2\mu}{\lambda} \left(1 + \frac{2\mu}{\lambda}\right)^{-(k+1)} \\ &= \frac{2\mu}{2\mu + \lambda} \left(1 - \frac{2\mu}{2\mu+\lambda}\right)^k, \quad k \in \{0, 1, 2, \ldots\}. \end{align}$$
When written this way, it becomes obvious that $X(T/2)$ is geometric with parameter $$\theta = \frac{2\mu}{2\mu + \lambda} = \frac{\mu}{\mu + \lambda/2}.$$