Solution:
$\frac{1}{2062300}$

We see that for 1 to be the last spot, 2 must be picked first (with probability $\frac{1}{n}$), after which spot $n$ is picked. Then, cars from 3 to $n-1$ will be picked until there are only gaps of 1 or 2 remaining. At this point, each of the remaining spots (including spot 1) is picked uniformly at random, so the probability that spot 1 is chosen last here will be the reciprocal of the number of remaining slots.

Let $f(n)$ denote the number of empty spots that will be left if cars park in $n+2$ consecutive spots whose ends are occupied, under the same conditions, except that the process stops when a car is forced to park immediately next to a car. We want to find the value of $f(2009)$. Given the gap of $n$ cars, after placing a car, there are gaps of $f\left(\left\lfloor\frac{n-1}{2}\right\rfloor\right)$ and $f\left(\left\lceil\frac{n-1}{2}\right\rceil\right)$ remaining. Thus, $f(n) = f\left(\left\lfloor\frac{n-1}{2}\right\rfloor\right) + f\left(\left\lceil\frac{n-1}{2}\right\rceil\right)$. With the base cases $f(1) = 1$, $f(2) = 2$, we can determine with induction that

$$f(x) = \begin{cases} x - 2^{n-1} + 1 & \text{if } 2^n \leq x \leq \frac{3}{2} \cdot 2^n - 2, \\ 2^n & \text{if } \frac{3}{2} \cdot 2^n - 1 \leq x \leq 2 \cdot 2^n - 1. \end{cases}$$

Thus, $f(2009) = 1024$, so the total probability is $\frac{1}{2012} \cdot \frac{1}{1024+1} = \frac{1}{2062300}$.