Use the Remainder Theorem for the first condition.

The polynomial is $f(x) = 3 - ax + bx^2 - x^3$.
When $f(x)$ is divided by $(x+2)$, the remainder is 45. By the Remainder Theorem, $f(-2) = 45$.
Substitute $x = -2$ into $f(x)$:

$$f(-2) = 3 - a(-2) + b(-2)^2 - (-2)^3$$ $$45 = 3 + 2a + 4b - (-8)$$ $$45 = 3 + 2a + 4b + 8$$ $$45 = 11 + 2a + 4b$$ $$34 = 2a + 4b$$

Divide the equation by 2:

$$17 = a + 2b \quad (*)$$

$(3-x)$ is a factor of $f(x)$. By the Factor Theorem, if $(3-x)$ is a factor, then $f(3) = 0$.
Substitute $x = 3$ into $f(x)$:

$$f(3) = 3 - a(3) + b(3)^2 - (3)^3$$ $$0 = 3 - 3a + 9b - 27$$ $$0 = -24 - 3a + 9b$$

Rearrange the equation:

$$24 = -3a + 9b$$

Divide the equation by 3:

$$8 = -a + 3b \quad (**)$$

We have two equations:

1. $a + 2b = 17$
2. $-a + 3b = 8$

Add equation (1) and equation (2):

$$(a + 2b) + (-a + 3b) = 17 + 8$$ $$5b = 25$$ $$b = \frac{25}{5}$$ $$b = 5$$

Substitute $b=5$ into equation (1):

$$a + 2(5) = 17$$ $$a + 10 = 17$$ $$a = 17 - 10$$ $$a = 7$$

The values of the real constants are $a=7$ and $b=5$.

$$\boxed{a=7, b=5}$$

ii) Express $\frac{5x}{(x+3)(x-2)}$ into partial fractions.

Since the denominator has two distinct linear factors, $(x+3)$ and $(x-2)$, we can write:

$$\frac{5x}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}$$

Multiply both sides by $(x+3)(x-2)$:

$$5x = A(x-2) + B(x+3)$$

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

$$5(-3) = A(-3-2) + B(-3+3)$$ $$-15 = A(-5) + B(0)$$ $$-15 = -5A$$ $$A = \frac{-15}{-5}$$ $$A = 3$$

To find $B$, set $x = 2$:

$$5(2) = A(2-2) + B(2+3)$$ $$10 = A(0) + B(5)$$ $$10 = 5B$$ $$B = \frac{10}{5}$$ $$B = 2$$

Substitute the values of $A$ and $B$ back into the decomposition:

$$\frac{5x}{(x+3)(x-2)} = \frac{3}{x+3} + \frac{2}{x-2}$$ $$\boxed{\frac{3}{x+3} + \frac{2}{x-2}}$$