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
$$
F(t) = 1- e^{-\Lambda(0,t)}
$$
Let $t=T_F$. Conditional on the number of occurences $N=n$, the arrival times $t_1,t_2,\dots,t_N$ are known to have the same distribution as the order statstics of $n$ iid unif$(0,t)$ random variables. Hence, the likelihood becomes
\begin{align}
L(\lambda,t) &= P(N=n) f(t_1,t_2,\dots,t_N|N=n) \\
&= \frac{e^{-\lambda t}(\lambda t)^n}{n!}\frac{n!}{t^n} \\
&= e^{-\lambda t}\lambda^n.
\end{align}
for $t\ge t_n$ and zero elsewhere. This is maximised for $\hat t=t_n$ and $\hat\lambda=n/t_n$. These MLEs don't exist if there are no occurrences $N=0$, however. Conditional on $N=n$, again using the fact that $t_n$ can be viewed as an order statistic (the maximum) of $n$ iid unif$(0,t)$ random variables, $E(t_N|N=n)=\frac n{n+1} t$. Hence, the estimator $t^*=\frac {n+1}n t_n$ is unbiased for $t$ conditional on $N=n$ and hence also conditional on $N\ge 1$. A reasonable frequentist estimator of $\lambda$ might be $\lambda^* = n/t^* = \frac{n^2}{(n+1)t_n}$ but this does not have finite expectation when $N=1$ so assessing its bias is even more troublesome.
Bayesian inference using independent, non-informative scale priors on $\lambda$ and $t$ on the other hand leads to a posterior
$$
f(\lambda,t|t_1,\dots,t_N) \propto e^{-\lambda t}\lambda^{n-1}t^{-1}.
$$
for $t>t_n,\lambda>0$. Integrating out $\lambda$, the marginal posterior of $t$ becomes
$$
f(t|t_1,\dots,t_N) = \frac{n t_n^n}{t^{n+1}}, t>t_n,
$$
and the posterior mean $E(t|t_1,\dots,t_N)=\frac n{n-1} t_n$. A $(1-\alpha)$-credible interval for $t$ is given by $\left(\frac{t_n}{(1-\alpha/2)^{1/n}}, \frac{t_n}{(\alpha/2)^{1/n}}\right)$.
The marginal posterior of $\lambda$,
\begin{align}
f(\lambda|t_1,\dots,t_N) &\propto \int_{t_\text{max}}^\infty e^{-\lambda t}\lambda^{n-1}t^{-1} dt \\
&= \lambda^{n-1}\Gamma(0,\lambda t_n)
\end{align}
where $\Gamma$ is the incomplete gamma function.
Best Answer
You can use Maximum Likelihood Estimation, either with synchronous data (time-binned data) or asynchronous data (time-stamped data). The likelihood function changes accordingly.
For time-binned (or synchronous) data, you can simply use the joint Poisson probability mass function for your observed counts as the likelihood function:
$ L = \prod^{K}_{i=1} \frac{\lambda^{x_i}}{x_i!} \exp(-\lambda) $,
where $K$ is the number of bins, $x_i$ the count of events in bin $i$, and $\lambda$ the constant intensity that you want to estimate.
For asynchronous data, the likelihood is specified as follows:
$L = \left[ \prod^{N(T)}_{i=1} \lambda^*(t_i) \right] \exp\left[-\int^{T}_{0}\lambda^*(s) ds \right] $,
where $N(T)$ is the number of points at end-of-sample time $T$, and $\lambda^*(t)$ is the conditional intensity function, which is simply the constant $\lambda^*(t)=\lambda$ for the homogeneous Poisson process.
In some cases including the homogeneous Poisson process, there are closed-form solutions for both cases (take logs, set derivative with respect to $\lambda$ equal to zero, and solve for $\lambda$). Otherwise the log-likelihood can be optimised numerically.
For more background on theory and estimation, these are good references: