Working with p-adic integers is really simple, and usually if you want to do rational numbers, it's fairly easy; you just use long-division.
But here in the case 9/10 when we're working in the 5-adics; I have no idea. I understand you can extend the p-adic integers to the p-adic numbers; but given $9 = 4 + 1 \cdot 5 + 0 \cdot 5^2 + ...$ and $10 = 0 + 2 \cdot 5 + 0 \cdot 5^2 + ...$ we can't do long-division. So it seems that we'd have to get it in p-adic number form; the issue here is though that I have no idea how to do this. Is it simply trial in error? If so, the question after this is $\frac{1234}{625}$ how the hell am I meant to do that!!!!!
Note that $1/5$ (or any power of it) is not a $5$-adic integer. It is just $\ldots0000.1$, a $5$-adic rational.
Since the geometric series gives $\ldots1111=\sum_n5^n=1/(1-5)=-1/4$, we have
$$\ldots2222=-1/2$$ $$\ldots2223=(-1/2)+1=1/2$$
so
$$9/10=(9/2)/5=(4+1/2)/5=(\ldots2232)/5=\ldots2223.2$$
For the second question, the key is that $625=5^4$ is a power of $5$. So
$$1234/625=0.1234$$
In general, you can simply shift the radix point for each factor of $5$ in the denominator; and the remaining factors in the denominator must be factors of some $(5^m-1)$, so you can move them to the numerator by using the geometric series $\sum_n5^{mn}=1/(1-5^m)$.