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
$$
F(t) = 1- e^{-\Lambda(0,t)}
$$
You have a tradeoff between power (e.g. the ability to see as quickly as possible when your process is not following a homogeneous Poisson process) and flexibility (the ability to see that it was a Poisson process the whole time, but perhaps the rate changed at some point, making it non-homogenous).
Note that Poisson processes have several features; there's not just that you have an exponential inter-event time but that the rate is constant and events occur independently of each other.
If you want to test for a homogenous process, looking at the inter-event times makes sense, and in that case you would maximize power by combining all your inter-event times into one sample. You could take counts in many sub-intervals and test for Poisson-ness but I don't think that will give you more power than testing for an exponential-distributed inter-event time.
You could gain substantial power if you have a specific alternative in mind and tailor your test against that alternative. Failing being able to nominate a specific sort of alternative there are several possible tests for an exponential distribution. Indeed, D'Agostino & Stephens' book (Goodness of fit Techniques) has an entire chapter on tests for the exponential -- and it doesn't include smooth tests.
There's some additional specifics on testing (and diagnostics) for exponential distributions in this thread
However you might also consider looking for some time dependence or non-homogeneity that might not be so obvious in the marginal distribution of inter-event times.
Note that however you test it - whether with one test or several tests of different aspects of the process - failure to reject doesn't mean that you have a Poisson process. It simply means you couldn't detect that it wasn't.
That is, the answer to Q2 is "No". You can't verify that the inter-arrival times are exponentially distributed, nor can you verify that they are random. You may be able to see that they probably aren't but failure to demonstrate that doesn't mean that they are. Maybe the effect was just smaller than you could reliably detect at this sample size.
In practice you will almost never have a true Poisson process but may in many cases have something well-approximated by one. With very large samples you'll be very likely to reject even trivial deviations from a Poisson process, even when it would be suitable for every practical purpose you had in mind for it. [Both rejection and failure to reject may not tell you what you need to know.]
For this reason you may be better to look at diagnostic plots that would let you examine deviations from the exponential or Poisson as well as looking at time-dependence or changes in rate over time. These will focus more nearly on how much deviation there is from the ideal case rather than on significance, which will be more useful from a modelling point of view (i.e. it will come a bit closer to answering a question like "will this Poisson-process model be a suitable approximation for my present purpose?")
Best Answer
Memorylessness is a property of the following form:
$$\Pr(X>m+n \mid X > m)=\Pr(X>n)\ .$$
This property holds for $X_1=\ \text{time to the next event in a Poisson process}\ $, but it doesn't hold for $X_k=\ \text{time to the}\, k^\text{th}\, \text{event in a Poisson process}\ $ when $k>1$.
As for how to show it, you could try to do it from first principles.
If you can show that the essentially equivalent form $P(X>s+t)\neq P(X>s)P(X>t)$, (for $s, t>0\ $), that would be sufficient; you already know the distribution for $X_k$.