We have a list of mathematical problems to solve, including equations to solve, expressions to evaluate and simplify, and finding the square of a number.

1. Problem Description

We have a list of mathematical problems to solve, including equations to solve, expressions to evaluate and simplify, and finding the square of a number.

2. Solution Steps

Here are the steps to solve each problem:

Problem 1: Solve the equation $3^k = \frac{81^2 \times 3^5}{3^{11}}$

First, we can express 81 as $3^4$. So, $81^2 = (3^4)^2 = 3^8$.

Then, the equation becomes $3^k = \frac{3^8 \times 3^5}{3^{11}}$.

Using the exponent rule $a^m \times a^n = a^{m+n}$, we have $3^8 \times 3^5 = 3^{8+5} = 3^{13}$.

So, the equation is $3^k = \frac{3^{13}}{3^{11}}$.

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$, we have $\frac{3^{13}}{3^{11}} = 3^{13-11} = 3^2$.

Therefore, $3^k = 3^2$. Hence, $k = 2$.

Problem 2(a): Evaluate $\sqrt[3]{0.027}$

We can rewrite 0.027 as $\frac{27}{1000}$.

So, $\sqrt[3]{0.027} = \sqrt[3]{\frac{27}{1000}} = \frac{\sqrt[3]{27}}{\sqrt[3]{1000}} = \frac{3}{10} = 0.3$.

Problem 2(b): Evaluate $(\frac{1}{9})^{\frac{1}{2}}$

$(\frac{1}{9})^{\frac{1}{2}} = \sqrt{\frac{1}{9}} = \frac{\sqrt{1}}{\sqrt{9}} = \frac{1}{3}$.

Problem 2(c): Evaluate $3^0 \times 3^{-2}$

Using the exponent rule $a^0 = 1$, we have $3^0 = 1$.

Using the exponent rule $a^{-n} = \frac{1}{a^n}$, we have $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$.

Therefore, $3^0 \times 3^{-2} = 1 \times \frac{1}{9} = \frac{1}{9}$.

Problem 3: Solve the equation $4^{2n-3} = 8$

We can rewrite both sides with base

2. $4 = 2^2$, so $4^{2n-3} = (2^2)^{2n-3} = 2^{2(2n-3)} = 2^{4n-6}$.

$8 = 2^3$.

So, the equation becomes $2^{4n-6} = 2^3$.

Therefore, $4n-6 = 3$.

$4n = 9$, so $n = \frac{9}{4}$.

Problem 4(a): Evaluate $2^{1.6} \times 2^{0.4}$

Using the exponent rule $a^m \times a^n = a^{m+n}$, we have $2^{1.6} \times 2^{0.4} = 2^{1.6+0.4} = 2^2 = 4$.

Problem 4(b): Evaluate $(\frac{12}{27})^{-\frac{3}{2}}$

First simplify the fraction $\frac{12}{27} = \frac{4}{9}$.

So, $(\frac{12}{27})^{-\frac{3}{2}} = (\frac{4}{9})^{-\frac{3}{2}} = (\frac{9}{4})^{\frac{3}{2}} = (\sqrt{\frac{9}{4}})^3 = (\frac{3}{2})^3 = \frac{3^3}{2^3} = \frac{27}{8}$.

Problem 5(a): Simplify $(2+x)^0$

Using the rule $a^0 = 1$ for any non-zero $a$, $(2+x)^0 = 1$, assuming $x \neq -2$.

Problem 5(b): Simplify $\frac{3x^{-2}}{3x}$

$\frac{3x^{-2}}{3x} = \frac{3}{3} \times \frac{x^{-2}}{x} = 1 \times x^{-2-1} = x^{-3} = \frac{1}{x^3}$.

Problem 6: Find the square of $\frac{1}{4}$

$(\frac{1}{4})^2 = \frac{1^2}{4^2} = \frac{1}{16}$.

Problem 7(a): Evaluate $2^3 \times 2^0$

$2^3 = 8$ and $2^0 = 1$.

Therefore, $2^3 \times 2^0 = 8 \times 1 = 8$.

Problem 7(b): Evaluate $\sqrt{1.6 \times 10^2 + 9}$

$1.6 \times 10^2 = 1.6 \times 100 = 160$.

$\sqrt{1.6 \times 10^2 + 9} = \sqrt{160 + 9} = \sqrt{169} = 13$.

Problem 8: Find $\sqrt[3]{0.027}$

As solved previously in Problem 2(a), $\sqrt[3]{0.027} = 0.3$.

Problem 9: Simplify $(2a)^{-2} \times 3a^2$

$(2a)^{-2} = \frac{1}{(2a)^2} = \frac{1}{4a^2}$.

$(2a)^{-2} \times 3a^2 = \frac{1}{4a^2} \times 3a^2 = \frac{3a^2}{4a^2} = \frac{3}{4}$.

3. Final Answer

Problem 1: $k = 2$

Problem 2(a): $0.3$

Problem 2(b): $\frac{1}{3}$

Problem 2(c): $\frac{1}{9}$

Problem 3: $n = \frac{9}{4}$

Problem 4(a): $4$

Problem 4(b): $\frac{27}{8}$

Problem 5(a): $1$

Problem 5(b): $\frac{1}{x^3}$

Problem 6: $\frac{1}{16}$

Problem 7(a): $8$

Problem 7(b): $13$

Problem 8: $0.3$

Problem 9: $\frac{3}{4}$