5. Let the increasing function $f(x)=k x-2$. It is stipulated: given an initial value $a$ of a real number $x$, assign $x_{1}=f(a)$, if $x_{1} \leqslant$ 2012, then assign $x_{2}=f\left(x_{1}\right)$, and so on; if $x_{n-1} \leqslant 2012$, then assign $x_{n}=f\left(x_{n-1}\right)$, otherwise stop the assignment; if $x_{n}$ is obtained, it is said that the assignment has been made $n$ times $\left(n \in \mathbf{N}_{+}\right)$. It is known that after assigning $n$ times, the process stops. Then, when the real number $a$ exists, the range of values for the real number $k$ is $\qquad$
5. $\left(\frac{1007}{1006},+\infty\right)$.

Obviously, $k>0, x_{1}=k a-2$.
When $n \geqslant 2$, we have $x_{n}=k x_{n-1}-2$.
(1) If $k=1$, then $x_{n}-x_{n-1}=-2$, so $x_{n}=x_{1}-(n-1) \cdot 2 = k a - 2 - (n-1) \cdot 2$.
(2) If $k>0$, and $k \neq 1$, let
$$
\left(x_{n}-\lambda\right)=k\left(x_{n-1}-\lambda\right) \text {. }
$$

Then $x_{n}=k x_{n-1}+\lambda-k \lambda$.
Comparing $x_{n}=k x_{n-1}-2$, we get
$$
\lambda-k \lambda=-2 \Rightarrow \lambda=\frac{2}{k-1} \text {. }
$$

Therefore, the sequence $\left\{x_{n}-\lambda\right\}$ is a geometric sequence with the first term $x_{1}-\lambda=k a-2-\lambda$ and common ratio $k$.
By the general term formula of a geometric sequence, we have
$$
x_{n}-\frac{2}{k-1}=\left(k a-2-\frac{2}{k-1}\right) k^{n-1}=\left(a-\frac{2}{k-1}\right) k^{n} \text {. }
$$

Thus, $x_{n}=a k^{n}+\frac{2-2 k^{n}}{k-1}$.
From $x_{n-1} \leqslant 20120$, and $k \neq 1$, we get
$$
k>\frac{1007}{1006} \text {. }
$$

Combining (1) and (2), we have $k \in\left(\frac{1007}{1006},+\infty\right)$.