Simplify the expression (4x⁴y^(-2z)16x²y^(-1z⁻²))^(1)/(2).

First, simplify the fraction inside the parentheses:

$$\frac{4x^4y^{-2}z}{16x^2y^{-1}z^{-2}}$$

Simplify the coefficients:

$$\frac{4}{16} = \frac{1}{4}$$

Simplify the $x$ terms using the rule $\frac{a^m}{a^n} = a^{m-n}$:

$$\frac{x^4}{x^2} = x^{4-2} = x^2$$

Simplify the $y$ terms:

$$\frac{y^{-2}}{y^{-1}} = y^{-2 - (-1)} = y^{-2+1} = y^{-1}$$

Simplify the $z$ terms:

$$\frac{z^1}{z^{-2}} = z^{1 - (-2)} = z^{1+2} = z^3$$

Combine these simplified terms:

$$\frac{1}{4} x^2 y^{-1} z^3 = \frac{x^2 z^3}{4y}$$

Now, apply the exponent $\frac{1}{2}$ to the entire simplified fraction. Remember that raising to the power of $\frac{1}{2}$ is the same as taking the square root.

$$\left(\frac{x^2 z^3}{4y}\right)^{\frac{1}{2}} = \frac{(x^2)^{\frac{1}{2}} (z^3)^{\frac{1}{2}}}{(4)^{\frac{1}{2}} (y)^{\frac{1}{2}}}$$

Apply the power to each term:

$$(x^2)^{\frac{1}{2}} = x^{2 \times \frac{1}{2}} = x^1 = x$$ $$(z^3)^{\frac{1}{2}} = z^{\frac{3}{2}}$$ $$(4)^{\frac{1}{2}} = \sqrt{4} = 2$$ $$(y)^{\frac{1}{2}} = \sqrt{y}$$

Combine these results:

$$\frac{x z^{\frac{3}{2}}}{2 \sqrt{y}}$$ $$\boxed{\frac{x z^{\frac{3}{2}}}{2 \sqrt{y}}}$$

Recall that any non-zero number raised to the power of 0 is 1, so $x^0 = 1$.
Substitute $x^0 = 1$ into the expression:

$$-6(1) + 4x^2 + 3x^2$$ $$-6 + 4x^2 + 3x^2$$

Combine the like terms ($4x^2$ and $3x^2$):

$$-6 + (4+3)x^2$$ $$-6 + 7x^2$$

It is standard practice to write terms with higher powers first:

$$7x^2 - 6$$ $$\boxed{7x^2 - 6}$$

$3 \times 3 \times 3 \times 3 \times \frac{1}{3} \times \frac{1}{3}$.
Group the terms:

$$(3 \times 3 \times 3 \times 3) \times \left(\frac{1}{3} \times \frac{1}{3}\right)$$

Write the first group in index notation:

$$3 \times 3 \times 3 \times 3 = 3^4$$

Write the second group in index notation:

$$\frac{1}{3} \times \frac{1}{3} = \left(\frac{1}{3}\right)^2$$

Recall that $\frac{1}{a^n} = a^{-n}$, so $\frac{1}{3} = 3^{-1}$.
Therefore, $\left(\frac{1}{3}\right)^2 = (3^{-1})^2 = 3^{-1 \times 2} = 3^{-2}$.
Now multiply the two parts:

$$3^4 \times 3^{-2}$$

Use the rule $a^m \times a^n = a^{m+n}$:

$$3^{4 + (-2)} = 3^{4-2} = 3^2$$ $$\boxed{3^2}$$

Recall that any non-zero number raised to the power of 0 is 1, so $13^0 = 1$.
Substitute $13^0 = 1$ into the expression:

$$\left(\frac{1}{5}\right)^2$$

Apply the exponent to both the numerator and the denominator:

$$\frac{1^2}{5^2}$$

Calculate the squares:

$$\frac{1 \times 1}{5 \times 5} = \frac{1}{25}$$ $$\boxed{\frac{1}{25}}$$