The curve $C_1$ has equation $y = \frac{ax^2+bx+3}{x+2}$, where $a$ and $b$ are constants. It is given that $C_1$ has an asymptote $y = x - 3$.

1. State the value of $a$ by comparing the leading term for the oblique asymptote.

The oblique asymptote $y = mx + c$ is given by the quotient of the polynomial division. The leading term of the quotient is $\frac{ax^2}{x} = ax$.
Given the asymptote is $y = x - 3$, we compare the $x$ terms to get $ax = x$, so:
$a = 1$

2. Set up an identity using the given asymptote and an unknown remainder.

Since $y = x - 3$ is the oblique asymptote, we can express the rational function in terms of its quotient and a remainder $R$:
$\frac{x^2 + bx + 3}{x + 2} \equiv x - 3 + \frac{R}{x + 2}$

Multiplying throughout by $(x + 2)$:
$x^2 + bx + 3 \equiv (x - 3)(x + 2) + R$
$x^2 + bx + 3 \equiv x^2 - x - 6 + R$

3. Compare coefficients of $x$ to show that $b = -1$.

By comparing the coefficients of $x$ on both sides of the identity:
$b = -1$ (shown)

1. Substitute $a=1$ and $b=-1$ from part (a) and rearrange the curve equation into a quadratic equation in $x$.

Using the values $a=1$ and $b=-1$ from part (a):
$y = \frac{x^2 - x + 3}{x + 2}$
Multiply both sides by $(x + 2)$:
$y(x + 2) = x^2 - x + 3$
$yx + 2y = x^2 - x + 3$
Rearranging into a quadratic equation in terms of $x$:
$x^2 - (y + 1)x + (3 - 2y) = 0$

2. Apply the discriminant condition for real roots to formulate a quadratic inequality in $y$.

For $x$ to have real values (and hence for $y$ to be achievable on the curve), the discriminant of the quadratic equation must be non-negative ($\Delta \ge 0$).
$[-(y + 1)]^2 - 4(1)(3 - 2y) \ge 0$
$y^2 + 2y + 1 - 12 + 8y \ge 0$
$y^2 + 10y - 11 \ge 0$

3. Solve the quadratic inequality to find the set of values $y$ can take, and deduce the values it cannot take.

Factorising the quadratic expression:
$(y + 11)(y - 1) \ge 0$
Solving this inequality gives the set of values that $y$ can take:
$y \le -11 \quad \text{or} \quad y \ge 1$

Therefore, the set of values that $y$ cannot take is strictly between these values:
$-11 < y < 1$

1. Express the $y$-coordinates of $P$ and $Q$ in terms of the parameter $x$.

Since the locus is traced out as $x$ varies, $P$ and $Q$ share the same variable $x$-coordinate, so $x_1 = x_2 = x$.
From part (a), $a=1$ and $b=-1$, so the $y$-coordinate of $P$ on $C_1$ is $y_1 = \frac{x^2 - x + 3}{x + 2}$.
The $y$-coordinate of $Q$ on $C_2$ is $y_2 = \frac{3x - x^2}{x + 2}$.

2. Find the coordinates of the midpoint and simplify to find the cartesian equation.

Let the midpoint be $(X, Y)$.
The $x$-coordinate of the midpoint is $X = \frac{x + x}{2} = x$.
The $y$-coordinate of the midpoint is $Y = \frac{y_1 + y_2}{2}$.
$Y = \frac{1}{2} \left( \frac{x^2 - x + 3}{x + 2} + \frac{3x - x^2}{x + 2} \right)$
$Y = \frac{1}{2} \left( \frac{x^2 - x + 3 + 3x - x^2}{x + 2} \right) = \frac{1}{2} \left( \frac{2x + 3}{x + 2} \right)$
Since $X = x$, we replace $X$ with $x$ and $Y$ with $y$ to form the cartesian equation:
$y = \frac{2x + 3}{2x + 4}$