(1) Let $a = b = c = \frac{1}{3}$, $d = e = 0$, it is easy to see that, when arranging them around a circle, we can always get two neighboring numbers of $\frac{1}{3}$, and their product is $\frac{1}{9}$.

(2) For any five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$, without loss of generality, we assume that $a \ge b \ge c \ge d \ge e \ge 0$. Arrange these numbers around a circle in such a way as seen in the figure:

Since $a + b + c + d + e = 1$, we have $a + 3d \le 1$, and
$$a \cdot 3d \le \left(\frac{a + 3d}{2}\right)^2 \le \frac{1}{4}$$
then $ad \le \frac{1}{12}$.

Furthermore, $a + b + c \le 1$, then $b + c \le 1 - a \le 1 - \frac{b + c}{2}$, i.e. $b + c \le \frac{2}{3}$. So
$$bc \le \frac{(b + c)^2}{4} \le \frac{1}{9}$$
Since $ce \le ae \le ad$ and $bd \le bc$, then any neighboring numbers in this arrangement have their product less than $\frac{1}{9}$.