Solve the following problems: (a) Find x in the proportion p3 + pq + q2 : x :: p3 - q3 : (p-q)2. (b) Resolve into partial fraction: (3x+7)/((x+3)(x2+4)).

Substitute this into the expression for $x$.

$$x = \frac{(p^2 + pq + q^2)(p-q)^2}{(p-q)(p^2 + pq + q^2)}$$

$$x = \frac{(p-q)^2}{p-q}$$ $$x = p-q$$

The value of $x$ is $\boxed{p-q}$.

6. (b) Resolve into partial fraction: $\frac{3x+7}{(x+3)(x^2+4)}$

Since $(x^2+4)$ is an irreducible quadratic factor, its numerator will be a linear expression.

$$\frac{3x+7}{(x+3)(x^2+4)} = \frac{A}{x+3} + \frac{Bx+C}{x^2+4}$$

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

$$3(-3)+7 = A((-3)^2+4) + (B(-3)+C)(-3+3)$$ $$-9+7 = A(9+4) + 0$$ $$-2 = 13A$$ $$A = -\frac{2}{13}$$

Expand the equation:

$$3x+7 = Ax^2+4A + Bx^2+3Bx+Cx+3C$$ $$3x+7 = (A+B)x^2 + (3B+C)x + (4A+3C)$$

Compare coefficients:
Coefficient of $x^2$: $0 = A+B$
Since $A = -\frac{2}{13}$, then $0 = -\frac{2}{13} + B \implies B = \frac{2}{13}$.

Coefficient of $x$: $3 = 3B+C$
Substitute $B = \frac{2}{13}$:

$$3 = 3\left(\frac{2}{13}\right) + C$$ $$3 = \frac{6}{13} + C$$ $$C = 3 - \frac{6}{13} = \frac{39-6}{13} = \frac{33}{13}$$

(Alternatively, check constant term: $7 = 4A+3C$.
$7 = 4(-\frac{2}{13}) + 3(\frac{33}{13})$
$7 = -\frac{8}{13} + \frac{99}{13}$
$7 = \frac{91}{13}$
$7 = 7$. This confirms the values of $A$, $B$, and $C$.)

$$\frac{3x+7}{(x+3)(x^2+4)} = \frac{-\frac{2}{13}}{x+3} + \frac{\frac{2}{13}x+\frac{33}{13}}{x^2+4}$$ $$\frac{3x+7}{(x+3)(x^2+4)} = -\frac{2}{13(x+3)} + \frac{2x+33}{13(x^2+4)}$$

The partial fraction decomposition is $\boxed{-\frac{2}{13(x+3)} + \frac{2x+33}{13(x^2+4)}}$.