Rewriting the numeator and applying the quotient of powers property:

\(\displaystyle \frac{1}{1.25 \times 10^{-6}}\)

\(\displaystyle = \frac{1 \times 10^{0}}{1.25 \times 10^{-6}}\)

\(\displaystyle = \frac{1 }{1.25 } \times \frac{ 10^{0}}{ 10^{-6}}\)

\(\displaystyle =0.8 \times 10^{0- (-6)}\)

\(\displaystyle =0.8 \times 10^{6}\)

This is not in scientific notation, so we adjust it as follows, applying the product of powers property:

\(\displaystyle 0.8 \times 10^{6}\)

\(\displaystyle =8 \times 10^{-1} \times 10^{6}\)

\(\displaystyle =8 \times 10^{-1+6}\)

\(\displaystyle =8 \times 10^{5}\)

Any expression raised to the first power is equal to that expression, and any expression raised to the power of 0 is equal to 1, so

\(\displaystyle 4 \left ( 6xyz \right ) ^{1} - \left (24xyz \right )^{0}\)

\(\displaystyle = 4 \left ( 6xyz \right ) -1\)

\(\displaystyle = 24xyz -1\)

Any nonzero expression to the power of zero is equal to 1:

\(\displaystyle 6 (xyz)^{0} + 4(xyz)^{0} = 6 \times 1 + 4 \times 1 = 6 + 4 = 10\)

Rewriting the numerator and applying the quotient of powers property:

\(\displaystyle \frac{1}{2.5 \times 10 ^{5}}\)

\(\displaystyle = \frac{1\times 10 ^{0}}{2.5 \times 10 ^{5}}\)

\(\displaystyle = \frac{1 }{2.5 } \times \frac{ 10 ^{0}}{ 10 ^{5}}\)

\(\displaystyle = 0.4 \times 10 ^{0-5}\)

\(\displaystyle = 0.4 \times 10 ^{ -5}\)

Since this is not in scientific notation, adjust as follows:

\(\displaystyle 0.4 \times 10 ^{ -5}\)

\(\displaystyle = 4\times 10 ^{ -1} \times 10 ^{ -5}\)

\(\displaystyle = 4\times 10 ^{ -1 + (-5)}\)

\(\displaystyle = 4\times 10 ^{ -6}\)

To solve \(\displaystyle \frac{6\cdot5^{9}}{2\cdot 5^{7}}\), 6 should be divided by 3. The exponent will be equal to the exponent of the numerator minus the exponent of the denominator. This results in:

\(\displaystyle 3\cdot5^{2}\)

\(\displaystyle 3\cdot25\)

\(\displaystyle 75\)

\(\displaystyle 4^{3}\) is equal to \(\displaystyle 4\cdot4\cdot4\)

\(\displaystyle 4\cdot4=16\)

Therefore, \(\displaystyle 4\cdot4\cdot4=16\cdot4=64\)

Thus, 64 is the correct answer. 

\(\displaystyle 2^{5}\) is equal to \(\displaystyle 2\cdot2\cdot2\cdot2\cdot2\)

Given that \(\displaystyle 2\cdot2=4\), the above expression can be simplified to:

\(\displaystyle 4\cdot4\cdot2=16\cdot2=32\)

Therefore, 32 is the correct answer. 

\(\displaystyle 3^{3}\) is equal to \(\displaystyle 3\cdot3\cdot3\). 

Given that \(\displaystyle 3\cdot3=9\), it follows that 

\(\displaystyle 3\cdot3\cdot3=9\cdot3=27\). 

Therefore, \(\displaystyle 3^{3}\) is the correct answer. 

The bases of all three terms are alike.  Since the terms are of a specific power, the rule of exponents state that the powers can be added if the terms are multiplied.

\(\displaystyle 3^{-2}\times3^{-6}\times3^{6}= 3^{-2-6+6}= 3^{-2}\)

When we have a negative exponent, we we put the number and the exponent as the denominator, over \(\displaystyle 1\)

\(\displaystyle \frac{1}{3^2}=\frac{1}{9}\)

\(\displaystyle \frac{8x ^{-2}}{(3x)^{2}}\)

To solve this problem, we start with the parentheses and exponents in the denominator. 

\(\displaystyle = \frac{8x ^{-2}}{3^{2}x^{2}} = \frac{8x ^{-2}}{9x^{2}}\)

Next, we can bring the \(\displaystyle x^2\) from the denominator up to the numerator by making the exponent negative. 

\(\displaystyle = \frac{8x ^{-2-2}}{9} = \frac{8x ^{-4}}{9}\)

Finally, to get rid of the negative exponent we can bring it back down to the denominator. 

\(\displaystyle = \frac{8}{9x ^{4}}\)