Given Z1 = 2 + 3i and Z2 = 2 − i, express racZ1Z2 in the form a + bi and find the modulus and argument of Z1 imes Z2.

The conjugate of $2 - i$ is $2 + i$.

$$\frac{Z_1}{Z_2} = \frac{2 + 3i}{2 - i} \times \frac{2 + i}{2 + i}$$

$$(2 + 3i)(2 + i) = 2(2) + 2(i) + 3i(2) + 3i(i)$$ $$= 4 + 2i + 6i + 3i^2$$

Since $i^2 = -1$:

$$= 4 + 8i - 3$$ $$= 1 + 8i$$

$$(2 - i)(2 + i) = 2^2 - i^2$$ $$= 4 - (-1)$$ $$= 4 + 1$$ $$= 5$$

$$\frac{Z_1}{Z_2} = \frac{1 + 8i}{5} = \frac{1}{5} + \frac{8}{5}i$$

The expression in the form $a+bi$ is $\boxed{\frac{1}{5} + \frac{8}{5}i}$.

$$Z_1 \times Z_2 = (2 + 3i)(2 - i)$$ $$= 2(2) + 2(-i) + 3i(2) + 3i(-i)$$ $$= 4 - 2i + 6i - 3i^2$$

Since $i^2 = -1$:

$$= 4 + 4i - 3(-1)$$ $$= 4 + 4i + 3$$ $$= 7 + 4i$$

For a complex number $z = x + yi$, the modulus is $|z| = \sqrt{x^2 + y^2}$.

$$|Z_1 \times Z_2| = |7 + 4i| = \sqrt{7^2 + 4^2}$$ $$= \sqrt{49 + 16}$$ $$= \sqrt{65}$$

For $z = x + yi$ in the first quadrant ($x>0, y>0$), the argument is $\arg(z) = \arctan\left(\frac{y}{x}\right)$.
Since $7 + 4i$ is in the first quadrant:

$$\arg(Z_1 \times Z_2) = \arctan\left(\frac{4}{7}\right)$$

The modulus is $\boxed{\sqrt{65}}$ and the argument is $\boxed{\arctan\left(\frac{4}{7}\right)}$.

Given $|Z_3| = 5$ and $\arg(Z_3) = \frac{\pi}{3}$.

$$Z_3 = 5 \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)$$

Let $Z_{12} = Z_1 \times Z_2$.
$|Z_{12}| = \sqrt{65}$ and $\arg(Z_{12}) = \arctan\left(\frac{4}{7}\right)$.

$$Z_{12} = \sqrt{65} \left(\cos\left(\arctan\left(\frac{4}{7}\right)\right) + i \sin\left(\arctan\left(\frac{4}{7}\right)\right)\right)$$

If $z_a = r_a(\cos \theta_a + i \sin \theta_a)$ and $z_b = r_b(\cos \theta_b + i \sin \theta_b)$, then $\frac{z_a}{z_b} = \frac{r_a}{r_b}(\cos(\theta_a - \theta_b) + i \sin(\theta_a - \theta_b))$.
Here, $r_a = \sqrt{65}$, $\theta_a = \arctan\left(\frac{4}{7}\right)$, $r_b = 5$, $\theta_b = \frac{\pi}{3}$.

$$\frac{Z_1 \times Z_2}{Z_3} = \frac{\sqrt{65}}{5} \left(\cos\left(\arctan\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) + i \sin\left(\arctan\left(\frac{4}{7}\right) - \frac{\pi}{3}\right)\right)$$

The expression in polar form is $\boxed{\frac{\sqrt{65}}{5} \left(\cos\left(\arctan\left(\frac{4}{7}\right) - \frac{\pi}{3}\right) + i \sin\left(\arctan\left(\frac{4}{7}\right) - \frac{\pi}{3}\right)\right)}$.