For any nonzero $\displaystyle a$ and for any $\displaystyle m$, $\displaystyle a^{-m} = \frac{1}{a^{m}}$.

$\displaystyle 7^{-2} = \frac{1}{7^{2}} = \frac{1}{7 \times 7 } = \frac{1}{49}$

$\displaystyle 5 ^{-2} = \frac{1}{5 ^{2}} = \frac{1}{5 \times 5} = \frac{1}{25}$

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

Since $\displaystyle 25 < 49 < 64$, 

we can reverse the order when taking their reciprocals:

$\displaystyle \frac{1}{64} < \frac{1}{49}< \frac{1}{25}$

That is, $\displaystyle 2 ^{-6}< 7^{-2}< 5 ^{-2}$.

For any nonzero $\displaystyle a$ and for any $\displaystyle m$, $\displaystyle a^{-m} = \frac{1}{a^{m}}$.

Also, any negative number raised to the power of an odd number is equal to the opposite of the same power of its absolute value.

Combine these concepts:

$\displaystyle \left (-5 \right )^{-3} = \frac{1 }{\left (-5 \right )^{3}} = \frac{1}{- (5^{3})} = -\frac{1}{125}$

$\displaystyle \left (-4 \right )^{-3} = \frac{1 }{\left (-4 \right )^{3}} = \frac{1}{- (4^{3})} = -\frac{1}{64}$

$\displaystyle \left ( -2\right ) ^{-7} = \frac{1}{\left ( -2\right )^{7}} = \frac{1}{- \left ( 2 ^{7}\right )} = - \frac{1}{128}$

Since $\displaystyle 64< 125 < 128$,

the reciprocals are in reverse order of this:

$\displaystyle \frac{1}{64}> \frac{1}{125 }>\frac{1}{128}$

Their opposites reverse again:

$\displaystyle -\frac{1}{64}< - \frac{1}{125 }< -\frac{1}{128}$

Or, equivalently,

$\displaystyle \left (-4 \right )^{-3} < \left (-5 \right )^{-3} < \left ( -2\right ) ^{-7}$.

Evaluate each term first.

$\displaystyle 2^2 =4$

$\displaystyle 4^3 = 64$

Divide 4 with 64 and reduce.

$\displaystyle \frac{4}{64} = \frac{(4\times 1)}{(4\times 16)}$

The answer is:  $\displaystyle \frac{1}{16}$

Do not subtract the exponents or divide the bases!  We will need to compute the numerator and denominator first.

$\displaystyle \frac{2^3}{6^2} = \frac{(2)(2)(2)}{(6)(6)} = \frac{8}{36}$

Reduce this fraction.

The answer is:  $\displaystyle \frac{2}{9}$

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

$\displaystyle x^a \div x^b = x^{a-b}$

So, we get

$\displaystyle m^6 \div m^2 = m^{6-2}$

$\displaystyle m^6 \div m^2 = m^4$

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

$\displaystyle x^a \cdot x^b = x^{a+b}$

So we get

$\displaystyle h^7 \cdot h^2 = h^{7+2}$

$\displaystyle h^7 \cdot h^2 = h^9$

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

$\displaystyle x^a \div x^b = x^{a-b}$

Now, let’s divide.  We get

$\displaystyle a^{12} \div a^4 = a^{12-4}$

$\displaystyle a^{12} \div a^4 = a^8$

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

$\displaystyle x^a \cdot x^b = x^{a+b}$

Now, we will multiply. We get

$\displaystyle x^9 \cdot x^2 = x^{9 + 2}$

$\displaystyle x^9 \cdot x^2 = x^{11}$

Simplify the numerator.

$\displaystyle (-2)^3=(-2)(-2)(-2) = -8$

Simplify the denominator.

$\displaystyle (-4)^{-1} = \frac{1}{(-4)^1} =-\frac{1}{4}$

Divide the terms.

$\displaystyle \frac{(-2)^3}{(-4)^{-1}} = \frac{-8}{-\frac{1}{4}} = -8\div- \frac{1}{4}$

Take the reciprocal of the second term and change the sign to a multiplication.

$\displaystyle -8\div- \frac{1}{4} = -8\times -4 = 32$

The answer is:  $\displaystyle 32$

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

$\displaystyle x^a \div x^b = x^{a-b}$

So, we get

$\displaystyle n^8 \div n^2 = n^{8-2}$

$\displaystyle n^8 \div n^2 = n^6$