To multiply variables with exponents, we will use the following formula:

\(\displaystyle x^a \cdot x^b = x^{a+b}\)

So, given the problem

\(\displaystyle h^3 \cdot h^9\)

we can solve. We get

\(\displaystyle h^3 \cdot h^9 = h^{3+9}\)

\(\displaystyle h^3 \cdot h^9 = h^{12}\)

\(\displaystyle (-4)^{5}\) has 5, an odd number, as an exponent, so \(\displaystyle (-4)^{5}\) is a negative number, and it can be calculated by taking the opposite of the fifth power of 4.

\(\displaystyle -4^{5}\) is equal to the opposite of the fifth power of 4 as well. 

Therefore,

\(\displaystyle (-4)^{5} = -4^{5} = - (4^{5})\).

\(\displaystyle 4^{5} = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\)

\(\displaystyle = 16 \cdot 4 \cdot 4 \cdot 4\)

\(\displaystyle = 64\cdot 4 \cdot 4\)

\(\displaystyle =256 \cdot 4\)

\(\displaystyle = 1,024\),

so

\(\displaystyle (-4)^{5} = -4^{5} = - 1,024\).

\(\displaystyle (-3)^{6}\) has 6, an even number, as an exponent, so \(\displaystyle (-3)^{6}\) is a positive number, and it can be calculated by taking sixth power of 3.

\(\displaystyle -3^{6}\) is the (negative) opposite of the sixth power of 3.

Therefore, 

\(\displaystyle (-3)^{6} = 3 ^{6} = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3\)

\(\displaystyle = 9 \cdot 3 \cdot 3 \cdot 3 \cdot 3\)

\(\displaystyle = 27 \cdot 3 \cdot 3 \cdot 3\)

\(\displaystyle = 81 \cdot 3 \cdot 3\)

\(\displaystyle =243 \cdot 3\)

\(\displaystyle = 729\)

and

\(\displaystyle -3^{6} = - (3^{6})= -729\).

In order to solve, we will need to eliminate the negative exponent.  Use the following property of negative exponents.

\(\displaystyle x^{-n} = \frac{1}{x^n}\)

\(\displaystyle (\frac{2}{3}) ^{-2} = \frac{1}{(\frac{2}{3}) ^{2}} = \frac{1}{(\frac{2}{3})(\frac{2}{3})}\)

Simplify the denominator.

\(\displaystyle \frac{1}{\frac{4}{9}} = 1\div\frac{4}{9}\)

Convert the division sign to multiplication and take the reciprocal of the second term.

\(\displaystyle 1\div\frac{4}{9} = 1\times \frac{9}{4}\)

The answer is:  \(\displaystyle \frac{9}{4}\)

To multiply variables with exponents, we will use the following formula:

\(\displaystyle x^a \cdot x^b = x^{a+b}\)

Now, let’s combine the following:

\(\displaystyle x^3 \cdot x^5 = x^{3+5}\)

\(\displaystyle x^3 \cdot x^5 = x^8\)

To divide variables with exponents, we will use the following formula:

\(\displaystyle \frac{x^a}{x^b} = x^{a-b}\)

So, we get

\(\displaystyle \frac{m^{10}}{m^2} = m^{10-2}\)

\(\displaystyle \frac{m^{10}}{m^2} = m^8\)

Evaluate the second term first.

\(\displaystyle (2^2)^3 = (2^2)(2^2)(2^2) = (4)(4)(4) = 64\)

Replace this term.

\(\displaystyle 3-(2^2)^3 = 3-64 = -61\)

The answer is:  \(\displaystyle -61\)

Evaluate each term using order of operations.  We can expand the terms in parentheses to eliminate the exponents.

\(\displaystyle (-1)^3-(2^2)^4 = (-1)(-1)(-1) - (2^2)(2^2)(2^2)(2^2)\)

Simplify the terms.

\(\displaystyle =(-1)(-1)(-1) - (2^2)(2^2)(2^2)(2^2)\)

\(\displaystyle = (-1)(-1)(-1) -(4)(4)(4)(4)\)

\(\displaystyle =-1 -256 = -257\)

The answer is:  \(\displaystyle -257\)

To multiply like terms with exponents, we will use the following formula:

\(\displaystyle a^x \cdot a^y = a^{x+y}\)

Now, given the problem

\(\displaystyle s^7 \cdot s^2\)

we can solve. We get

\(\displaystyle s^7 \cdot s^2 = s^{7+2}\)

\(\displaystyle s^7 \cdot s^2 = s^9\)

In order to solve, we can simplify by multiplying the powers together according to the rule of exponents.

\(\displaystyle (a^{X})^Y = a^{XY}\)

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

The negative exponent can be rewritten as a fraction.

\(\displaystyle 3^{-6} = \frac{1}{3^6} = \frac{1}{729}\)

The answer is:  \(\displaystyle \frac{1}{729}\)