How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.

I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem: 
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok. 
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$

For $n=4$ and $k=5$ here is a solution:
$\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$
$\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$
$\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$
$\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$
$\left(4-\frac{4}{\left(4-\left(4+4\right)\right)}\right)=5$
$\left(4+\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=6$
$\frac{4}{\left(4 \times \frac{4}{\left(4+4!\right)}\right)}=7$
$\left(4 \times \left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=8$
$\left(4-\left(\frac{4}{4}-\frac{4!}{4}\right)\right)=9$
$\left(4+\frac{4}{\left(4 \times \frac{4}{4!}\right)}\right)=10$
$\frac{4}{\left(4 \times \frac{4}{44}\right)}=11$
$\left(4-\left(4-\left(4+\left(4+4\right)\right)\right)\right)=12$
