(1) For any $a \in A$, $a = 2^k$ ($k \in \mathbb{N}^*$). Then $2a = 2^{k+1}$. Let $b$ be any positive integer strictly less than $2a - 1$. Then $(b+1) \le 2a - 1$.

Between $b$ and $b+1$, one is an odd number that contains no prime factor $2$, and the other is an even number that contains at most the $k$th power of $2$. Therefore, $b(b+1)$ is definitely not a multiple of $2a$.

(2) For $a \in \bar{A}$ and $a \ne 1$, suppose $a = 2^k m$ where $k$ is a non-negative integer and $m$ is an odd number greater than $1$. Then $2a = 2^{k+1} m$. We will present three different proofs in the following.

Proof 1. Let $b = mx$, $b+1 = 2^{k+1} y$. Eliminating $b$, we have $2^{k+1}y - mx = 1$. Since $(2^{k+1}, m) = 1$, the equation has integral solutions that can be expressed as
$$\begin{cases} x = x_0 + 2^{k+1} t, \\ y = y_0 + m t \end{cases} \quad (\text{where } t \in \mathbb{Z}, \text{ and } (x_0, y_0) \text{ is a special solution of the equation}.)$$
Denote the smallest solution among them as $(x^*, y^*)$. Then $x^* < 2^{k+1}$.
Therefore, $b = m x^* < 2a - 1$ and $b(b+1)$ is a multiple of $2a$.

Proof 2. Since $(2^{k+1}, m) = 1$, by the Chinese Remainder Theorem, the congruence equation
$$\begin{cases} x \equiv 0 \pmod{2^{k+1}}, \\ x \equiv m-1 \pmod{m} \end{cases}$$
has a solution $x = b$ with $b \in (0, 2^{k+1} m)$. It is easy to see that $b < 2a-1$ and $b(b+1)$ is a multiple of $2a$.

Proof 3. Since $(2^{k+1}, m) = 1$, then there exists $r \in \mathbb{N}^*$, $r \le m - 1$, such that $2^r \equiv 1 \pmod{m}$.
Take $t \in \mathbb{N}^*$ such that $tr > k + 1$. Then $2^{tr} \equiv 1 \pmod{m}$. It is easy to see that there exists
$$b = (2^{tr} - 1) - q \cdot 2^{k+1} m > 0 \quad (q \in \mathbb{N}),$$
such that $0 < b < 2a - 1$. Then we have $m \mid b$, $2^{k+1} \mid b + 1$.
Therefore, $b(b + 1)$ is a multiple of $2a$.