Given the equation:

\[
5 \sqrt{5} \times 5^3 \div 5^{-3 / 2} = 5^{(a+2)}
\]

We start by expressing all terms as powers of 5:

\[
5^1 \times 5^{\frac{1}{2}} \times 5^3 \div 5^{-\frac{3}{2}} = 5^{a+2}
\]

When you combine the exponents, you add them:

\[
5^{1 + \frac{1}{2} + 3} \times 5^{\frac{3}{2}} = 5^{a+2}
\]

Since multiplying with the same base allows you to add exponents:

\[
5^{\frac{2}{2} + \frac{1}{2} + \frac{6}{2} + \frac{3}{2}} = 5^{a+2}
\]

Simplify the exponents:

\[
5^{\frac{12}{2}} = 5^{a+2}
\]

Which simplifies further to:

\[
5^6 = 5^{a+2}
\]

Setting the exponents equal to each other gives us:

\[
a+2 = 6
\]

Solving for \(a\):

\[
a = 6 – 2
\]
\[
a = 4
\]

Therefore, the value of \(a\) is \(\boldsymbol{4}\).

Given the operation \(x^{*} y = x^{2} + y^{2} – xy\), let’s calculate the value of \(9^{*} 11\).

Substitute \(x = 9\) and \(y = 11\) into the formula:

\[
9^{*} 11 = 9^{2} + 11^{2} – 9 \times 11
\]

Simplify each term:

\[
9^{*} 11 = 81 + 121 – 99
\]

Add and subtract the terms:

\[
9^{*} 11 = 202 – 99 = 103
\]

Therefore, the value of \(9^{*} 11\) is \(\boldsymbol{103}\).

Given \(a^b = 125\) and knowing \(a\) and \(b\) are positive integers, let’s first identify \(a\) and \(b\).

Finding \(a\) and \(b\)

The number 125 is a perfect cube, specifically \(5^3\). Therefore, \(a = 5\) and \(b = 3\).

Calculating \((a-b)^{a+b-4}\)

Now, substitute \(a = 5\) and \(b = 3\) into the expression:

\[
(a-b)^{a+b-4} = (5-3)^{5+3-4}
\]

This simplifies to:

\[
(2)^{4} = 16
\]

Therefore, the value of \((a-b)^{a+b-4}\) is \(\boldsymbol{16}\).

Identification of the Unique Bus Number

The task is to identify a bus number within the range of 1 to 500 that is not only a perfect square but also remains a perfect square when the digits are viewed upside down. Following the clarified criteria:

Criteria for the Bus Number

• The number must be a perfect square.
• When turned upside down, the number must still represent a perfect square.
• The number must include only the digits 0, 1, 6, 8, and 9, as these are the digits that can represent other numbers or themselves when flipped.

Analysis

We consider the range of perfect squares from \(1^2\) to \(22^2\) as these are the perfect squares within the 1 to 500 range:

\[
1, 4, 9, \ldots, 484\ (22^2)
\]

Among these, the number that meets the specific requirement of being a perfect square that, when turned upside down, also becomes a perfect square, is 169:

– Original Number: \(169 = 13^2\), a perfect square.
– Upside Down: When 169 is flipped upside down, it becomes 961. Notably, 6 and 9 flip, while 1 remains the same, making the new figure, 961, which is \(31^2\), another perfect square.

Conclusion

Therefore, the bus number with the described peculiarity is 169. This number is unique in that it satisfies the condition of being a perfect square and also transforms into another perfect square (961) when viewed upside down.

To simplify the given expression, we can use the conjugate of each denominator to rationalize it. The conjugate of a binomial \(\sqrt{a} – \sqrt{b}\) is \(\sqrt{a} + \sqrt{b}\). Multiplying both the numerator and denominator by the conjugate, we get:

\[
\left[\frac{1}{\sqrt{9}-\sqrt{8}}\right] – \left[\frac{1}{\sqrt{8}-\sqrt{7}}\right] + \left[\frac{1}{\sqrt{7}-\sqrt{6}}\right] – \left[\frac{1}{\sqrt{6}-\sqrt{5}}\right] + \left[\frac{1}{\sqrt{5}-\sqrt{4}}\right]
\]

\[
= \left[\frac{\sqrt{9} + \sqrt{8}}{(\sqrt{9} – \sqrt{8})(\sqrt{9} + \sqrt{8})}\right] – \left[\frac{\sqrt{8} + \sqrt{7}}{(\sqrt{8} – \sqrt{7})(\sqrt{8} + \sqrt{7})}\right] + \left[\frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7} – \sqrt{6})(\sqrt{7} + \sqrt{6})}\right]
\]

\[
– \left[\frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6} – \sqrt{5})(\sqrt{6} + \sqrt{5})}\right] + \left[\frac{\sqrt{5} + \sqrt{4}}{(\sqrt{5} – \sqrt{4})(\sqrt{5} + \sqrt{4})}\right]
\]

\[
= \left[\frac{\sqrt{9} + \sqrt{8}}{9 – 8}\right] – \left[\frac{\sqrt{8} + \sqrt{7}}{8 – 7}\right] + \left[\frac{\sqrt{7} + \sqrt{6}}{7 – 6}\right] – \left[\frac{\sqrt{6} + \sqrt{5}}{6 – 5}\right] + \left[\frac{\sqrt{5} + \sqrt{4}}{5 – 4}\right]
\]

\[
= (\sqrt{9} + \sqrt{8}) – (\sqrt{8} + \sqrt{7}) + (\sqrt{7} + \sqrt{6}) – (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + \sqrt{4})
\]

Now, notice that the terms cancel out in pairs:

\[
= \sqrt{9} + \sqrt{4} = 3 + 2 = 5
\]

Therefore, the value of the given expression is 5.