Determine z1 × z2 and give the answer in polar form. Determine z1 + z2 and give the answer in polar form.

Recall that $i^2 = -1$ and $i^4 = 1$.
The given equation is:

$$5x^2i - 4 = 3yi^4 + y^2i^2 - y(xi + i)$$

Substitute the values of $i^4$ and $i^2$:

$$5x^2i - 4 = 3y(1) + y^2(-1) - yxi - yi$$ $$5x^2i - 4 = 3y - y^2 - yxi - yi$$

Left side:

$$-4 + 5x^2i$$

Right side:

$$(3y - y^2) + (-yx - y)i$$

Equate the real parts:

$$-4 = 3y - y^2$$

Equate the imaginary parts:

$$5x^2 = -yx - y$$

$$-4 = 3y - y^2$$ $$y^2 - 3y - 4 = 0$$

Factor the quadratic equation:

$$(y - 4)(y + 1) = 0$$

This gives two possible values for $y$: $y = 4$ or $y = -1$.
The problem states that $y \in \mathbb{N}$, which means $y$ must be a natural number (a positive integer).
Therefore, we choose the positive value for $y$.

$$y = 4$$

$$5x^2 = -yx - y$$

Substitute $y = 4$:

$$5x^2 = -4x - 4$$

Rearrange into a standard quadratic form:

$$5x^2 + 4x + 4 = 0$$

To solve for $x$, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Here, $a=5$, $b=4$, $c=4$.

$$x = \frac{-4 \pm \sqrt{4^2 - 4(5)(4)}}{2(5)}$$ $$x = \frac{-4 \pm \sqrt{16 - 80}}{10}$$ $$x = \frac{-4 \pm \sqrt{-64}}{10}$$

Since $\sqrt{-64} = \sqrt{64 \times -1} = 8i$:

$$x = \frac{-4 \pm 8i}{10}$$

Simplify the expression:

$$x = \frac{-2 \pm 4i}{5}$$

The solutions are:
$y = \boxed{\text{4}}$
$x = \boxed{-\frac{2}{5} + \frac{4}{5}i \text{ or } -\frac{2}{5} - \frac{4}{5}i}$