Observe first that
$$\binom{2n+1}{n} = \frac{(2n+1)((2n)!)}{n!(n+1)!} = (2n+1)C_n,$$
so
$$C_n = (2n + 2 - (2n + 1))C_n = 2 \binom{2n}{n} - \binom{2n+1}{n},$$
so $C_n$ is an integer, as all binomial coefficients $\binom{m}{k}$ with positive integers $m, k$ with $k \le m$ are integers. This establishes (i).

To prove (ii), note that
$$C_{n+1} = \frac{1}{n+2} \binom{2n+2}{n+1} = \frac{(2n+2)(2n+1)((2n)!)}{(n+2) \cdot ((n+1)!)^2} = \frac{2(2n+1)}{n+2} C_n,$$
hence
$$(n + 2)C_{n+1} = 2(2n + 1)C_n. \qquad (1)$$
By definition
$$\begin{aligned} C_n &= \frac{1}{n+1} \binom{2n}{n} = \frac{2n(2n-1)\cdots(n+1)}{(n+1)n(n-1)\cdots2 \cdot 1} \\ &= \frac{(2n-1)(2n-2)\cdots(n+3)(n+2)}{(n-1)\cdots3} \end{aligned}$$

Thus, for $n > 3$,
$$\frac{n+2}{C_n} = \frac{(n-1)(n-2)\cdots3}{(2n-1)(2n-2)\cdots(n+3)} < 1,$$
since each factor $h$ in the numerator is matched by a factor $h+n$ in the denominator. Thus $C_n$ does not divide $n+2$.