Given the expression (1)/(124) x² - A + (1)/(4) is a perfect square.

Since $p^2 = \frac{1}{124}$, $p$ cannot be 0.
From $2pq = 0$, and knowing $p \neq 0$, it must be that $q = 0$.

Substitute $q=0$ into the constant term equation:
$q^2 = -A + \frac{1}{4}$
$0^2 = -A + \frac{1}{4}$
$0 = -A + \frac{1}{4}$

$A = \frac{1}{4}$

The value of $A$ is $\boxed{\frac{1}{4}}$.

:
Given the equation $\log_3 (x^2-9) = 3\log_2 2 + 1$.

Recall that $\log_b b = 1$. Therefore, $\log_2 2 = 1$.
RHS $= 3(1) + 1 = 3 + 1 = 4$.

$\log_3 (x^2-9) = 4$.

If $\log_b y = z$, then $y = b^z$.
So, $x^2-9 = 3^4$.

$3^4 = 3 \times 3 \times 3 \times 3 = 81$.

$x^2 - 9 = 81$
$x^2 = 81 + 9$
$x^2 = 90$
$x = \pm\sqrt{90}$

$\sqrt{90} = \sqrt{9 \times 10} = \sqrt{9} \times \sqrt{10} = 3\sqrt{10}$.

So, $x = \pm 3\sqrt{10}$.

For $\log_3 (x^2-9)$ to be defined, the argument must be positive: $x^2-9 > 0$.
This means $x^2 > 9$.
If $x = 3\sqrt{10}$, then $x^2 = (3\sqrt{10})^2 = 9 \times 10 = 90$. Since $90 > 9$, this solution is valid.
If $x = -3\sqrt{10}$, then $x^2 = (-3\sqrt{10})^2 = 9 \times 10 = 90$. Since $90 > 9$, this solution is also valid.

The values of $x$ are $\boxed{\pm 3\sqrt{10}}$.