The Poisson Distribution Customers arrive at a casino as a Poisson process of rate 100 customers per hour. Upon
arriving, each customer must flip a coin, and only those customers who flips heads actually enter the casino. 

So how can I find


the PMF of N? Let's say for the
number of customers who arrive
between 5PM to 7PM. 
the pdf of T? The    time between
successive customers    arriving.
The key observation is that the new process, denote it by $N = \lbrace N(t) : t \geq 0 \rbrace$, has rate $\lambda$ which is half the original, that is $\lambda=50$.
Then the answers to your questions follow immediately. Denote the original process by $N_0$, and set $\lambda_0 = 100$.
Then, the distribution of $N(1)$ (that is, the number of customers who actually enter the casino in one hour) is obtained (using the law of total probability) as follows.
$$
{\rm P}(N(1) = k) = \sum\limits_{j = k}^\infty  {{\rm P}(N(1) = k|N_0 (1) = j) {\rm P}(N_0 (1) = j)}.
$$
Given $N_0 (1) = j$, $N(1)$ is binomial$(j,1/2)$. Hence,
$$
{\rm P}(N (1) = k) = \sum\limits_{j = k}^\infty  {\frac{{e^{ - \lambda _0 } \lambda _0^j }}{{j!}}{j \choose k}\frac{1}{{2^j }}} = \cdots = \frac{{e^{ - \lambda _0 /2} (\lambda _0 /2)^k }}{{k!}}.
$$

EDIT: For a proof that the new process $N$ indeed has stationary independent increments, see, for example, p. 211 here (where $\lambda$ and $p$ correspond to $100$ and $1/2$, respectively).
Note that the pdf of $T$ is an exponential distribution.