For a real number α , let [α] denote the greatest integer less than...

For the function $f(x) = [x^3]\log_e(1 + \sin^2(\pi(x - [x])))$, we can simplify the argument of the sine function. Since $[x]$ is an integer, $\sin(\pi(x - [x])) = \sin(\pi x - \pi[x]) = \pm \sin(\pi x)$. Thus, $\sin^2(\pi(x - [x])) = \sin^2(\pi x)$.

The function $f(x)$ can be rewritten as $f(x) = [x^3]\log_e(1 + \sin^2(\pi x))$.
The term $\log_e(1 + \sin^2(\pi x))$ is continuous everywhere. The term $[x^3]$ has jump discontinuities where $x^3$ is an integer. For $x \in (-3, 3)$, $x^3 \in (-27, 27)$. The integer values of $x^3$ in this interval are $k \in \{-26, -25, \dots, 26\}$, which gives $53$ points of the form $x = k^{1/3}$.

At these points, $[x^3]$ is discontinuous. For $f(x)$ to be continuous at $x = k^{1/3}$, the continuous factor must be zero:
$\log_e(1 + \sin^2(\pi x)) = 0 \implies \sin^2(\pi x) = 0 \implies x$ is an integer.
The integers in $(-3, 3)$ are $-2, -1, 0, 1, 2$. Their cubes are $-8, -1, 0, 1, 8$, which are $5$ values among the $53$ points. At these $5$ points, $f(x)$ is continuous. At the remaining $53 - 5 = 48$ points, $f(x)$ is discontinuous.
Thus, $|A| = 48$.

For the function $g(x) = x^3 \sin^2(\pi \log_e(1 + x - [x]))$, we can write $x - [x] = \{x\}$, which is the fractional part of $x$.
The function $g(x) = x^3 \sin^2(\pi \log_e(1 + \{x\}))$ is continuous everywhere except possibly at the integers, where $\{x\}$ is discontinuous.
The integers in $(-3, 3)$ are $c \in \{-2, -1, 0, 1, 2\}$.
Let's check the continuity at $x = c$:
Right-hand limit ($x \to c^+$): $\{x\} \to 0$, so $g(x) \to c^3 \sin^2(\pi \log_e 1) = 0$.
Left-hand limit ($x \to c^-$): $\{x\} \to 1$, so $g(x) \to c^3 \sin^2(\pi \log_e 2)$.

For $g(x)$ to be continuous at $x = c$, we must have $c^3 \sin^2(\pi \log_e 2) = 0$. Since $\log_e 2 \approx 0.693$, $\pi \log_e 2$ is not an integer multiple of $\pi$, meaning $\sin^2(\pi \log_e 2) \neq 0$. Therefore, we must have $c^3 = 0 \implies c = 0$.
So, $g(x)$ is continuous at $x = 0$ and discontinuous at $x \in \{-2, -1, 1, 2\}$.
Thus, $B = \{-2, -1, 1, 2\}$ and $|B| = 4$.

Since $A$ contains no integers and $B$ contains only integers, $A \cap B = \emptyset$, which means $|A \cap B| = 0$.

Finally, we calculate the required value:
$|A| + 2|B| - |A \cap B| = 48 + 2(4) - 0 = 48 + 8 = 56$.

Answer: $56$