Split (x2)/((x2+9)(x-3)) into partial fractions.

The denominator has an irreducible quadratic factor $(x^2+9)$ and a linear factor $(x-3)$.

$$\frac{x^2}{(x^2+9)(x-3)} = \frac{Ax+B}{x^2+9} + \frac{C}{x-3}$$

To find C, let $x=3$:

$$(3)^2 = (A(3)+B)(3-3) + C((3)^2+9)$$ $$9 = 0 + C(9+9)$$ $$9 = 18C$$ $$C = \frac{9}{18} = \frac{1}{2}$$

Expand the right side of the equation from Step 2:

$$x^2 = Ax^2 - 3Ax + Bx - 3B + Cx^2 + 9C$$

Group terms by powers of $x$:

$$x^2 = (A+C)x^2 + (-3A+B)x + (-3B+9C)$$

Equate coefficients:

1. Coefficient of $x^2$: $A+C = 1$
2. Coefficient of $x$: $-3A+B = 0$
3. Constant term: $-3B+9C = 0$

Substitute $C=\frac{1}{2}$ into equation (1):

$$A + \frac{1}{2} = 1$$ $$A = 1 - \frac{1}{2} = \frac{1}{2}$$

Substitute $A=\frac{1}{2}$ into equation (2):

$$-3\left(\frac{1}{2}\right) + B = 0$$ $$-\frac{3}{2} + B = 0$$ $$B = \frac{3}{2}$$

(As a check, substitute B and C into equation (3): $-3\left(\frac{3}{2}\right) + 9\left(\frac{1}{2}\right) = -\frac{9}{2} + \frac{9}{2} = 0$. This is correct.)

$$\frac{x^2}{(x^2+9)(x-3)} = \frac{\frac{1}{2}x+\frac{3}{2}}{x^2+9} + \frac{\frac{1}{2}}{x-3}$$ $$= \frac{x+3}{2(x^2+9)} + \frac{1}{2(x-3)}$$

The partial fraction decomposition is $\boxed{\frac{x+3}{2(x^2+9)} + \frac{1}{2(x-3)}}$.

The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+4)$.

$$\frac{1}{(x-1)(x^2+4)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+4}$$

To find A, let $x=1$:

$$1 = A((1)^2+4) + (B(1)+C)(1-1)$$ $$1 = A(1+4) + 0$$ $$1 = 5A$$ $$A = \frac{1}{5}$$

Expand the right side of the equation from Step 2:

$$1 = Ax^2 + 4A + Bx^2 - Bx + Cx - C$$

Group terms by powers of $x$:

$$1 = (A+B)x^2 + (-B+C)x + (4A-C)$$

Equate coefficients:

1. Coefficient of $x^2$: $A+B = 0$
2. Coefficient of $x$: $-B+C = 0$
3. Constant term: $4A-C = 1$

Substitute $A=\frac{1}{5}$ into equation (1):

$$\frac{1}{5} + B = 0$$ $$B = -\frac{1}{5}$$

Substitute $B=-\frac{1}{5}$ into equation (2):

$$-(-\frac{1}{5}) + C = 0$$ $$\frac{1}{5} + C = 0$$ $$C = -\frac{1}{5}$$

(As a check, substitute A and C into equation (3): $4\left(\frac{1}{5}\right) - \left(-\frac{1}{5}\right) = \frac{4}{5} + \frac{1}{5} = \frac{5}{5} = 1$. This is correct.)

$$\frac{1}{(x-1)(x^2+4)} = \frac{\frac{1}{5}}{x-1} + \frac{-\frac{1}{5}x-\frac{1}{5}}{x^2+4}$$ $$= \frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)}$$

The partial fraction decomposition is $\boxed{\frac{1}{5(x-1)} - \frac{x+1}{5(x^2+4)}}$.

To find A, let $x=-3$:

$$6(-3)+7 = A(-3+1) + B(-3+3)$$ $$-18+7 = A(-2) + 0$$ $$-11 = -2A$$ $$A = \frac{-11}{-2} = \frac{11}{2}$$

To find B, let $x=-1$:

$$6(-1)+7 = A(-1+1) + B(-1+3)$$ $$-6+7 = 0 + B(2)$$ $$1 = 2B$$ $$B = \frac{1}{2}$$

$$B-A = \frac{1}{2} - \frac{11}{2}$$ $$B-A = \frac{1-11}{2}$$ $$B-A = \frac{-10}{2}$$ $$B-A = -5$$

The value of $(B-A)$ is $\boxed{-5}$.