Question on effective rate of interest given rate continuous compounding $\mathbf{\text{(i) Given r= $0.12 $ annually compounded, find the monthly}}$
 $\mathbf{\text{repayments on a $ 1 000 000$ Dollar loan to be repaid completely in 5 years}}$ 
$\mathbf{\text{Solution}}$
Finding equivalent rate of monthly compounding:
$(1+ 0.12)= (1+\frac{1_{12}}{12})^{12} \rightarrow r_{12}=0.11387$
Then 
$1 000 000= C [\frac{1-(1+\frac{0.1139}{12})^{-60}}{\frac{0.1139}{12}}] \rightarrow C= 21 937.42$
$\mathbf{\text{(ii) Given r= $0.12 $ with continuous compounding, find the monthly}}$ $\mathbf{\text{ repayments on a $ 1 000 000$ Dollar loan to be repaid completely in 5 years}}$ 
$e ^ {-0.12}=(1+\frac{r_{12}}{12})^{12}$
Been getting $r_{12}$ as negative. Please help. 

The factor for an continuously compounding is $e^{r}$, not $e^{-r}$
There is a well known identity:
$$\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n=e^x$$
Since in your case n is not very large the equation doesn´t hold. But you can say that the left side is monthly compounding (with n=12) and the right side is continuously compounding. The x on both sides has to be different if the equation should be true.
$\left(1+\frac{r_{12}}{12}\right)^{12}=e^{0.12}$
Taking the 12-th root on both sides.
$\left(1+\frac{r_{12}}{12}\right)^{1}=e^{0.01}$
$\frac{r_{12}}{12}=e^{0.01}-1$
$r_{12}=12\cdot (e^{0.01}-1)=0.120602...$
