You have to pay attention to the directions of the arrows you have drawn.
For the nodes, it usually does not matter which direction you draw the arrows, though it helps to have at least one arrow going in and make sure you follow the currents around since current out of a component = current into a component.
Take a look at what you've drawn for node A.
On your diagram, I8 and i9 go out while I7 is in, so you write
I7=I8+I9 ... correctly.
Follow I8 around the circuit - see that I2 = I8? For similar reasons, I1 = I7 and so on - which is why the diagram that comes with the problem has so few currents on it.
So - use the current labelled on the diagram you are given.
When you get to the voltage law - you have six loops total, though you can probably get away with the obvious three. Draw arrows for each power source (batteries here) pointing from negative to positive and label with the voltage of each (there's only one here and it's done for you) but also draw arrows on all the resistors - these arrows point opposite the currents and label them IR (with the right numbers/subscripts for the I and the R).
Start at a node, and trace around one loop. Each time you go against an arrow, you subtract it, each time you go along an arrow you add it, when you get back to the start you write "=0".
This all will give you a set of simultaneous equations.