Let Z be the set of integers. If $$A = \{x \in Z : 2^{(x+2)(x^2-5x+6)} = 1\}$$ and $$B = \{x \in Z : -3 < 2x - 1 < 9\}$$, then the number of subsets of the set $$A \times B$$, is:

We have the set $$A=\{x\in \mathbb Z:2^{(x+2)\bigl(x^{2}-5x+6\bigr)}=1\}.$$

First, recall the basic fact: for any real number $$k,$$ the equality $$2^{k}=1$$ holds only when $$k=0,$$ because the base $$2$$ is positive and not equal to $$1.$$ Hence, to find all integers $$x$$ belonging to $$A,$$ we must force the exponent to be zero:

$$ (x+2)\bigl(x^{2}-5x+6\bigr)=0. $$

This product equals zero precisely when at least one factor is zero. We split it:

1. $$x+2=0 \;\Longrightarrow\; x=-2.$$

2. $$x^{2}-5x+6=0.$$

We factor the quadratic:

$$ x^{2}-5x+6=(x-2)(x-3). $$

So

$$ (x-2)(x-3)=0 \;\Longrightarrow\; x=2 \text{ or } x=3. $$

Collecting all integer solutions, we obtain

$$ A=\{-2,\,2,\,3\}, $$

which contains $$|A|=3$$ elements.

Now consider the set $$B=\{x\in\mathbb Z:-3<2x-1<9\}.$$ We solve the double inequality step by step.

First, add $$1$$ to every part:

$$ -3+1 < 2x-1+1 < 9+1 \;\Longrightarrow\; -2 < 2x < 10. $$

Next, divide every part by $$2$$ (the sign of $$2$$ is positive, so the inequalities keep their direction):

$$ \frac{-2}{2} < x < \frac{10}{2} \;\Longrightarrow\; -1 < x < 5. $$

The integers strictly between $$-1$$ and $$5$$ are

$$ 0,\;1,\;2,\;3,\;4. $$

Thus

$$ B=\{0,\,1,\,2,\,3,\,4\}, $$

with $$|B|=5$$ elements.

We now form the Cartesian product $$A\times B=\{(a,b):a\in A,\;b\in B\}.$$ The counting principle tells us that

$$ |A\times B|=|A|\times|B|=3\times5=15. $$

Finally, the total number of subsets of any finite set containing $$n$$ elements is given by the formula $$2^{n}.$$ Substituting $$n=15,$$ we get

$$ \text{Number of subsets}=2^{15}. $$

Hence, the correct answer is Option D.