Simplify the following exponential expressions: (a3)2, (b2)4, (c5)3, (d4)3, (e3)3, (f0)8, (g⁻²)5, (h4)⁻⁵.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(a^3)^2 = a^{3 \times 2} = a^6$$

The simplified expression is $\boxed{\mathbf{a^6}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(b^2)^4 = b^{2 \times 4} = b^8$$

The simplified expression is $\boxed{\mathbf{b^8}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(c^5)^3 = c^{5 \times 3} = c^{15}$$

The simplified expression is $\boxed{\mathbf{c^{15}}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(d^4)^3 = d^{4 \times 3} = d^{12}$$

The simplified expression is $\boxed{\mathbf{d^{12}}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(e^3)^3 = e^{3 \times 3} = e^9$$

The simplified expression is $\boxed{\mathbf{e^9}}$.

First, simplify $f^0$. Using the rule $a^0 = 1$:

$$f^0 = 1$$

Then, substitute this back into the expression:

$$(f^0)^8 = (1)^8 = 1$$

The simplified expression is $\boxed{\mathbf{1}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(g^{-2})^5 = g^{-2 \times 5} = g^{-10}$$

Using the rule $a^{-n} = \frac{1}{a^n}$:

$$g^{-10} = \frac{1}{g^{10}}$$

The simplified expression is $\boxed{\mathbf{\frac{1}{g^{10}}}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(h^4)^{-5} = h^{4 \times (-5)} = h^{-20}$$

Using the rule $a^{-n} = \frac{1}{a^n}$:

$$h^{-20} = \frac{1}{h^{20}}$$

The simplified expression is $\boxed{\mathbf{\frac{1}{h^{20}}}}$.

First, calculate $5^2$:

$$5^2 = 25$$

Then, substitute this back into the expression:

$$(5^2)^{-1} = (25)^{-1}$$

Using the rule $a^{-n} = \frac{1}{a^n}$:

$$(25)^{-1} = \frac{1}{25^1} = \frac{1}{25}$$

The simplified expression is $\boxed{\mathbf{\frac{1}{25}}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

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

Calculate the value:

$$3^4 = 3 \times 3 \times 3 \times 3 = 81$$

The simplified expression is $\boxed{\mathbf{81}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(10^2)^7 = 10^{2 \times 7} = 10^{14}$$

The simplified expression is $\boxed{\mathbf{10^{14}}}$.

Using the rule $(a^m)^n = a^{m \times n}$:

$$(2^{-3})^2 = 2^{-3 \times 2} = 2^{-6}$$

Using the rule $a^{-n} = \frac{1}{a^n}$:

$$2^{-6} = \frac{1}{2^6}$$

Calculate the value:

$$\frac{1}{2^6} = \frac{1}{2 \times 2 \times 2 \times 2 \times 2 \times 2} = \frac{1}{64}$$

The simplified expression is $\boxed{\mathbf{\frac{1}{64}}}$.