Given f(x) = ax4 + 7x3 + x2 + bx - 3, if (x-1) and (x+1) are factors, find a and b, and then find the set of real values of x for which f(x) > 0.

The polynomial is $f(x) = ax^4 + 7x^3 + x^2 + bx - 3$.
Given that $(x-1)$ and $(x+1)$ are factors of $f(x)$, we know that $f(1)=0$ and $f(-1)=0$.

For $f(1)=0$:

$$a(1)^4 + 7(1)^3 + (1)^2 + b(1) - 3 = 0$$ $$a + 7 + 1 + b - 3 = 0$$ $$a + b + 5 = 0$$ $$a + b = -5 \quad (*)$$

For $f(-1)=0$:

$$a(-1)^4 + 7(-1)^3 + (-1)^2 + b(-1) - 3 = 0$$ $$a(1) + 7(-1) + 1 - b - 3 = 0$$ $$a - 7 + 1 - b - 3 = 0$$ $$a - b - 9 = 0$$ $$a - b = 9 \quad (**)$$

We have:

1. $a + b = -5$
2. $a - b = 9$

Add equation (1) and equation (2):

$$(a+b) + (a-b) = -5 + 9$$ $$2a = 4$$ $$a = 2$$

Substitute $a=2$ into equation (1):

$$2 + b = -5$$ $$b = -5 - 2$$ $$b = -7$$

The values of the constants are $a=2$ and $b=-7$.

$$\boxed{a=2, b=-7}$$

Substitute $a=2$ and $b=-7$ into $f(x)$:

$$f(x) = 2x^4 + 7x^3 + x^2 - 7x - 3$$

Since $(x-1)$ and $(x+1)$ are factors, their product $(x-1)(x+1) = x^2-1$ is also a factor.
Perform polynomial long division to find the other factor:

$$(2x^4 + 7x^3 + x^2 - 7x - 3) \div (x^2 - 1)$$ $$\begin{array}{c|cc cc cc} \multicolumn{2}{r}{2x^2} & +7x & +3 \\ \cline{2-6} x^2-1 & 2x^4 & +7x^3 & +x^2 & -7x & -3 \\ \multicolumn{2}{r}{-(2x^4} & & -2x^2) \\ \cline{2-4} \multicolumn{2}{r}{0} & +7x^3 & +3x^2 & -7x \\ \multicolumn{2}{r}{} & -(7x^3 & & -7x) \\ \cline{3-5} \multicolumn{2}{r}{} & 0 & +3x^2 & 0 & -3 \\ \multicolumn{2}{r}{} & & -(3x^2 & & -3) \\ \cline{4-6} \multicolumn{2}{r}{} & & 0 & 0 & 0 \\ \end{array}$$

So, $f(x) = (x^2-1)(2x^2+7x+3)$.
Factor the quadratic term $2x^2+7x+3$:

$$2x^2+7x+3 = 2x^2+x+6x+3 = x(2x+1)+3(2x+1) = (x+3)(2x+1)$$

Thus, the fully factored form of $f(x)$ is:

$$f(x) = (x-1)(x+1)(x+3)(2x+1)$$

The roots of $f(x)$ are the values of $x$ where $f(x)=0$:
$x-1=0 \implies x=1$
$x+1=0 \implies x=-1$
$x+3=0 \implies x=-3$
$2x+1=0 \implies x = -\frac{1}{2}$

Order the roots on a number line: $-3, -1, -\frac{1}{2}, 1$.
These roots divide the number line into five intervals. We test the sign of $f(x)$ in each interval:

• For $x < -3$ (e.g., $x=-4$): $f(-4) = (-5)(-3)(-1)(-7) = 105 > 0$

• For $-3 < x < -1$ (e.g., $x=-2$): $f(-2) = (-3)(-1)(1)(-3) = -9 < 0$

• For $-1 < x < -\frac{1}{2}$ (e.g., $x=-0.75$): $f(-0.75) = (-1.75)(0.25)(2.25)(-0.5) \approx 0.49 > 0$

• For $-\frac{1}{2} < x < 1$ (e.g., $x=0$): $f(0) = (-1)(1)(3)(1) = -3 < 0$

• For $x > 1$ (e.g., $x=2$): $f(2) = (1)(3)(5)(5) = 75 > 0$

We are looking for $f(x) > 0$. This occurs in the intervals $(-\infty, -3)$, $(-1, -\frac{1}{2})$, and $(1, \infty)$.
The set of real values of $x$ for which $f(x) > 0$ is:

$$\boxed{x \in (-\infty, -3) \cup \left(-1, -\frac{1}{2}\right) \cup (1, \infty)}$$