current through 12 ohm is (6/12+4.5)
second part
this came from the voltage divider rule since 4.5 and 12 ohms are in series

This is still not right, because some of the current in R3 does not pass through R1,
so they are not technically in series.
If you were to cooperate then we would have finished this question by now and I like to think you would also be understanding what you are doing.
I asked for the total current which you are right in saying is given in my figure 7 as the total voltage divided by the R7, the equivalent resistance, but you did not calculate it.
But you need to be more precise in what you are saying
yes you just take the equivalent resistance divide by the voltage

The voltage?
Which voltage?
Note I said the total voltage (6 volts)
So the total current = 6/7.415 = 0.81 amps.
Now I also asked you
So once you have calculated the total current can you see what the current in R3 is?
One of the various stages I labelled 1 to 6 should help,Can you see which one?

So I labelled on of the junctions or nodes in figs 1 and 2 with an A.
This junction has 3 branches
The total current into the battery
The current into R4
The current into R3
So the current into R3 + Current into R4 must = Total current
Can you now calculate the current in R3?