For inhomogeneous Poisson process the probability that there are no jumps on the interval $[0,y]$ is
$$
\mathbb P(\tau >y) = e^{-\Lambda(y)}
$$
Recall that the distribution of $\tau$ is absolutely continuous and the pdf can be obtained from the above equality.
Since the function $\Lambda(t)$ is continuous and non-decreasing, we can define inverse of it as
$$
\Lambda^{-1}(x) = \inf\{t: \Lambda(t)\geq x\}.
$$
Note that $\Lambda(\tau)\geq x$ $\iff$ $\tau\geq\Lambda^{-1}(x)$.
Then
$$
\mathbb P(\Lambda(\tau)\geq x) = \mathbb P(\tau\geq\Lambda^{-1}(x))=\mathbb P(\tau >\Lambda^{-1}(x)) = e^{-\Lambda(\Lambda^{-1}(x))} = e^{-x}.
$$
So the distribution of $\xi=\Lambda(\tau)$ is standard exponential.
Answering your questions, in the first place the times are not independent unless the function $ \lambda $ is constant, that is, it is a homogeneous Poisson process. In the case of a non-homogeneous process, the value $ \lambda(u) $ represents the rate at time $ u $.
To find the density of $ T_{1} $, you just have to see that: $$1-P(T_{1}\leqslant t)=P(T_{1}>t)=P(N(t)=0)=e^{-\mu(t)}$$ where: $$\mu(t)=\int_{0}^{t}\lambda(u)\,du$$
Deriving this expression we obtain that:
$$f_{T_{1}}(t)=-\frac{\text{d}}{\text{d}t}P(T_{1}>t)=\lambda(t)e^{-\mu(t)}$$
and this is precisely the density of $ T_{1} $. This is exponential in the case where $ \lambda(u) = \text{const}$. If you wanted to find the joint distribution for example between $ T_ {1} $ and $ T_ {2} $, just find the conditional, that is $P(T_{2}>t_{2}|T_{1}=t_{1})$. Now, since we are stopped at moments $ t_ {1} $, the set where $ T_ {2}> t_ {2} $ is equivalent to the number of events occurring at time $ t_ {1} $ is equal to the number of events at time $ t_ {1} + t_ {2} $, that is: $$P(T_{2}>t_{2}|T_{1}=t_{1})=P(N(t_{1}+t_{2})-N(t_{1})=0|T_{1}=t_{1})=P(N(t_{1}+t_{2})-N(t_{1})=0|N(t_{1})\geqslant 1)=P(N(t_{1}+t_{2})-N(t_{1})=0)=e^{-[\mu(t_{1}+t_{2})-\mu(t_{1})]}$$
Deriving with respect to $ t_ {2} $ we can see that: $$f_{T_{2}|T_{1}}(t_{2}|t_{1})=\lambda(t_{1}+t_{2})e^{-[\mu(t_{1}+t_{2})-\mu(t_{1})]}$$
So you already have an expression for the joint density, which is: $$f_{T_{1},T_{2}}(t_{1},t_{2})=\lambda(t_{1})\lambda(t_{1}+t_{2})e^{-\mu(t_{1}+t_{2})}$$
So they are not independent. Notice that we have used throughout the proof that $ N (t) $ has independent increments, and that $ N(t+s)-N(t) $ has a Poisson distribution with parameter $\mu(t+s)-\mu(t)$
Best Answer
Unfortunately, the $T_i$ are not a poisson process. One way of showing it is, for example, by considering the number $N$ of $T_i$ that falls into the interval $[0,1]$. It's supposed to be a Poisson rv if the $T_i$ are a poisson process. But, $$\mathbb{P}(N=0)=\mathbb{P}(\tau_1\geq 1)=e^{-\lambda_1}$$ and $$\mathbb{P}(N=1)=\mathbb{P}(\tau_1\leq 1~\mbox{and}~\tau_1+\tau_2\geq 1)=\left\{\begin{array}{l} \lambda_1 e^{-\lambda_1}~\mbox{if}~\lambda_1=\lambda_2 \\ \frac{\lambda_1 e^{-\lambda_2}}{\lambda_2-\lambda_1}\left(e^{\lambda_2-\lambda_1}-1\right)~\mbox{otherwise.}\end{array}\right.$$ So, $\lambda_1=\lambda_2$ and similarly we can show that all $\lambda_i$ have to be equal.
An instinctive way of seeing that $T_i$ is not a poisson process is imagining, for example, that $\lambda_1=1$ and $\lambda_i=10$ for all $i\geq 2$. If we know that $T_1<1$ then we know that the average number of $T_i$ that falls in $[1,2[$ is $10$ but if $T_1>1$ then this is not the case anymore, it will be lower. So, knowing what happens on the interval $[0,1[$ has an influence on what happens on the interval $[1,2[$. This is never the case for a Poisson point process.