Define a sequence of prior distributions, $\pi_n = Ga(\lambda|a_n,b_n)$, for the sequence $a_n = \alpha/n$ and $b_n = \beta/n$.
The Bayes estimator for this sequence is $\delta_n = (a_n+x)(b_n+1)$, and the integrated risk is
\begin{gather}
r_n = \int^\infty_0 (\lambda-\delta_n)^2 Poi(x|\lambda)Ga(\lambda|a_n,b_n)d\lambda
\end{gather}
which reduces to
\begin{gather}
r_n = \delta_n^2 - 2\delta_n E_n\lambda + E_n\lambda^2
\end{gather}
\begin{gather}
= \delta_n^2 - 2\delta_n E_n\lambda + V_n\lambda +(E_n\lambda)^2
\end{gather}
\begin{gather}
= \delta_n^2 - 2\delta_n^2 + \delta_n(b_n+1) + \delta_n^2
\end{gather}
\begin{gather}
= \delta_n(b_n+1).
\end{gather}
The limiting Bayes estimator is
\begin{gather}
\delta = \lim_{n\rightarrow\infty}\delta_n=\lim_{n\rightarrow\infty}(a_n+x)(b_n+1) = x
\end{gather}
and the limiting integrated risk is
\begin{gather}
r = \lim_{n\rightarrow\infty}\delta_n(b_n+1)=\lim_{n\rightarrow\infty}(a_n+x)(b_n+1)^2=x
\end{gather}
The limiting rule $\delta$ is Bayes with respect to the improper prior $\pi_\infty$. Because the integrated risk $r$ is constant for all $\lambda$, the estimator $\delta = x$ is minimax.
In general, the times $T_i$ are not independent and not identically distributed. So questions 1 and 2 are answered in the negative. It is not too difficult to find examples in which they are clearly dependent and/or clearly differently distributed. I leave such a demonstration for someone else but shall however try to derive the distribution of $T_1$.
First recall that if $\lambda(t)=\lambda$ is fixed (so in case of the usual homogeneous Poisson process), the probability that there is no point in an interval of length $s$ is $e^{-\lambda s}$. This is because we know that the number of points $N_s$ in an interval of length $s$ is Poisson distributed with parameter $\lambda s$ (constant rate $\times$ interval length):
$$
\text{Prob}[N_s = n] = e^{-\lambda s} \frac{(\lambda s)^n}{n!}
$$
so
$$
\text{Prob}[N_s = 0] = e^{-\lambda s}
$$
Secondly, consider the probability that no point occurs in the interval $[0,s]$ for the nonhomogeneous process. To obtain this probability, we are going to cut the interval in equal parts of length $\Delta$ and then let $\Delta\to 0$. Every part then becomes very (infinitesimally) small so that we can assume the rate $\lambda(t)$ is constant in that small part. If the rate is constant there (say equal to $\lambda$), the number of points in that part is Poisson distributed with parameter $\lambda \Delta$. So we have:
\begin{align*}
\text{Prob}\big[\text{no point in }[0,s]\big]
& = \lim_{\Delta\to 0}\text{Prob}\big[\text{no point in }[0,\Delta], \ldots, \text{no point in }[s-\Delta,s]\big]\\
& = \lim_{\Delta\to 0}\prod_{i=0}^{s/\Delta-1}\text{Prob}\big[\text{no point in }[i\Delta,(i+1)\Delta]\big]\\
& = \lim_{\Delta\to 0}\prod_{i=0}^{s/\Delta-1} e^{-\lambda(i\Delta)\Delta}\\
& = \lim_{\Delta\to 0} \exp \Big[ - \sum_{i=0}^{s/\Delta-1} \lambda(i\Delta)\Delta) \Big]\\
& = \exp \Big[ - \lim_{\Delta\to 0} \sum_{i=0}^{s/\Delta-1} \lambda(i\Delta)\Delta \Big]
= \exp \Big[ - \int_0^s \lambda(t) \text{d}t \Big]
= e^{-\Lambda(0,s)}
\end{align*}
because the last limit is nothing but a Riemann sum that is the area under the graph of $\lambda(t)$ from $t=0$ to $t=s$. For convenience, let us call $\Lambda(t_1,t_2) =\int_{t_1}^{t_2} \lambda(t)\text{d}t$.
Finally, consider the probability density $f(t)$ function of $T_1$:
\begin{align*}
f(t)\text{d}t & = \text{Prob}[ t \leqslant T_1 < t+\text{d}t ]\\
& = \text{Prob}\big[\text{no point in }[0,t]\big] \cdot \text{Prob}\big[\text{1 point in }[t,t+\text{d}t[\big]\\
& = e^{-\Lambda(0,t)} \cdot e^{-\lambda(t)\text{d}t}\lambda(t)\text{d}t
= e^{-\Lambda(0,t)} \lambda(t)\text{d}t
\end{align*}
so the density of $T_1$ is
$$
f(t) = e^{-\Lambda(0,t)} \lambda(t)
$$
and the distribution function $F(t)=\text{Prob}[T\leqslant t]$ follows by integration of $f(t)$ as
$$
F(t) = 1- e^{-\Lambda(0,t)}
$$
Best Answer
Don't hesitate to use WolframAlpha to get the sum of a series. Or do you need a mathematical proof ?
This gives $\exp(-2\lambda)I_0(2\lambda)$. The link to the documentation of the Bessel function $I_0$ is this one.
Actually the proof here would just mean the series representation of $I_0$.
If you want to use R to evaluate this Bessel function, you can do it with the help of the
gslpackage: