We apply the exponents first before simplifying the fractions.

\(\displaystyle (\frac{x}{y})^4*(\frac{y^2}{x^{-3}})^2=\frac{x^4}{y^4}*\frac{y^4}{x^{-6}}\) The \(\displaystyle y^4\) cancels and we have \(\displaystyle \frac{x^4}{x^{-6}}\). When dividing exponents, we subtract the exponents and keep the base the same.

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

When dividing exponents, we subtract the exponents and keep the base the same.

\(\displaystyle \frac{x^7}{x^8}=x^{7-8}=x^{-1}\) We know with negative exponents, it's expressed as one over the positive exponent.

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

Let's apply the exponents to the parentheses first and then simplify.

\(\displaystyle x^5*(\frac{5}{x^4})^2*\frac{x^8}{10}=x^5*\frac{25}{x^8}*\frac{x^8}{10}\) The \(\displaystyle x^8\) cancels and the numbers can be reduced by \(\displaystyle 5\). 

We finally get: \(\displaystyle \frac{5x^5}{2}\).

Let's rewrite this as just exponents. Remember we can breakup \(\displaystyle \frac{x^4}{y^7}=x^4*\frac{1}{y^7}\).\(\displaystyle \frac{x^4}{y^7}*x^{-8}=\frac{1}{y^7}*x^4*x^{-8}=y^{-7}*x^{-4}\)

When dividing exponents, we subtract the exponents and keep the base the same.

\(\displaystyle \frac{5^{2x}}{5^{2x+4}}=5^{2x-(2x+4)}=5^{2x-2x-4}=5^{-4}=\frac{1}{5^4}\) We know with negative exponents, it's expressed as one over the positive exponent.

When dividing exponents, we subtract the exponents and keep the base the same.

\(\displaystyle \frac{7^{8x+4}}{7^{7x-2}}=7^{8x+4-(7x-2)}=7^{8x+4-7x+2}=7^{x+6}\)

Although it seems like we can't simplify anything, we do know that \(\displaystyle a^2+2ab+b^2=(a+b)^2\). Therefore we have:

\(\displaystyle \frac{a+b}{a^2+2ab+b^2}=\frac{a+b}{(a+b)^2}\). Now we can divide the exponents to get \(\displaystyle (a+b)^{-1}=\frac{1}{a+b}\)

Let's apply the exponential operation before we simplify.

\(\displaystyle (\frac{x}{y})^{7n}*(\frac{x}{y^{-1}})^{2n}*\frac{y^{5n}}{x^{5n+2}}=\frac{x^{7n}}{y^{7n}}*x^{2n}y^{2n}*\frac{y^{5n}}{x^{5n+2}}\) In the numerator, the \(\displaystyle y^{2n}, y^{5n}\) become \(\displaystyle y^{7n}\) and cancels with the denominator in the left fraction.

We now have: \(\displaystyle x^{7n}*x^{2n}*\frac{1}{x^{5n+2}}\). By combining the top and applying the division rule of exponents, we get: 

\(\displaystyle x^{7n}*x^{2n}*\frac{1}{x^{5n+2}}=\frac{x^{9n}}{x^{5n+2}}=x^{9n-(5n+2)}=x^{9n-5n-2}=x^{4n-2}\)

Although the exponents have different bases, we know that \(\displaystyle 4=2^2\). Therefore we can rewrite as \(\displaystyle \frac{2^{12}}{4^6}=\frac{2^{12}}{(2^2)^6}=\frac{2^{12}}{2^{12}}=1\)

Although the bases are not the same, we know that \(\displaystyle 8=2^3, 4=2^2\). We will base our answers in base of \(\displaystyle 2\) since this is present in all the choices. Therefore:  \(\displaystyle \frac{2^{24}8^8}{4^{16}}=\frac{2^{24}(2^3)^8}{(2^2)^{16}}=\frac{2^{24}2^{24}}{2^{32}}\) Now we add the exponents and then subtract them.

\(\displaystyle \frac{2^{24}2^{24}}{2^{32}}=\frac{2^{48}}{2^{32}}=2^{16}\)