Let's solve the problems step by step:

01. $f(x) = 2x - 9$ and $g(x) = x^2 + 5x + 3$

We need to find the value of the sum of the absolute values of the roots of the equation $f(g(x)) = g(x)$.

$f(g(x)) = 2g(x) - 9$ $g(x) = x^2 + 5x + 3$

Substituting $g(x)$ into $f$:

$f(g(x)) = 2(x^2 + 5x + 3) - 9$ $f(g(x)) = 2x^2 + 10x + 6 - 9$ $f(g(x)) = 2x^2 + 10x - 3$

Set this equal to $g(x)$:

$2x^2 + 10x - 3 = x^2 + 5x + 3$ $x^2 + 5x - 6 = 0$

Solve the quadratic equation:

$x = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm 7}{2}$

Roots are:

$x_1 = 1 \quad \text{and} \quad x_2 = -6$

Sum of the absolute values of the roots:

$|1| + |-6| = 1 + 6 = 7$

02. $f(x) = 2x$ and $g(x) = 2 - x$

We need to find $x$ in the equation $f(g(x)) + g(f(x)) = f(f(x)) + g(g(x))$.

Calculate each function composition:

$f(g(x)) = f(2 - x) = 2(2 - x) = 4 - 2x$ $g(f(x)) = g(2x) = 2 - 2x$ $f(f(x)) = f(2x) = 2(2x) = 4x$ $g(g(x)) = g(2 - x) = 2 - (2 - x) = x$

Substitute into the equation:

$(4 - 2x) + (2 - 2x) = 4x + x$ $6 - 4x = 5x$ $6 = 9x$ $x = \frac{2}{3}$

03. $f(x) = 3 - 4x$ and $g(x) = 3x + m$

We need to find the value of the constant $m$ such that $f(g(x)) = g(f(x))$ for any real $x$.

Calculate each function composition:

$f(g(x)) = f(3x + m) = 3 - 4(3x + m) = 3 - 12x - 4m$ $g(f(x)) = g(3 - 4x) = 3(3 - 4x) + m = 9 - 12x + m$

Equate both compositions:

$3 - 12x - 4m = 9 - 12x + m$ $3 - 4m = 9 + m$ $3 - 9 = 4m + m$ $-6 = 5m$ $m = -\frac{6}{5}$

04. $f(x) = 2x^2 - x + 1$ and $g(x) = x - 2$

a) Calculate $f \circ g \circ f \circ g \circ f \circ g (3)$:

$g(3) = 3 - 2 = 1$ $f(1) = 2(1)^2 - 1 + 1 = 2 - 1 + 1 = 2$ $g(2) = 2 - 2 = 0$ $f(0) = 2(0)^2 - 0 + 1 = 1$ $g(1) = 1 - 2 = -1$ $f(-1) = 2(-1)^2 - (-1) + 1 = 2 + 1 + 1 = 4$

Therefore, $f \circ g \circ f \circ g \circ f \circ g (3) = 4$.

b) Find the real values of $x$ for which $f(g(x)) \leq 2 \cdot g(x)$:

$f(g(x)) = 2(x - 2)^2 - (x - 2) + 1$ $2(x^2 - 4x + 4) - x + 2 + 1$ $= 2x^2 - 8x + 8 - x + 2 + 1$ $= 2x^2 - 9x + 11$

Inequality:

$2x^2 - 9x + 11 \leq 2(x - 2)$ $2x^2 - 9x + 11 \leq 2x - 4$ $2x^2 - 11x + 15 \leq 0$

Solve the quadratic inequality:

$x = \frac{11 \pm \sqrt{121 - 120}}{4}$ $x = \frac{11 \pm 1}{4}$ $x_1 = 3 \quad \text{and} \quad x_2 = \frac{5}{2}$

The values of $x$ that satisfy the inequality are $\frac{5}{2} \leq x \leq 3$.

Would you like more details or have any questions? Here are some additional problems you might consider:

1. How do you solve quadratic equations by factoring?
2. What is the process for finding the derivative of a function?
3. How can you integrate a polynomial function?
4. What are the properties of logarithms?
5. How do you determine the convergence of a sequence?

Tip: Always verify your final answers by substituting them back into the original equations to ensure they satisfy all given conditions.