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).