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
$$
F(t) = 1- e^{-\Lambda(0,t)}
$$
