1) Winding direction doesn't matter as long as you keep it consistent. If you are adapting a lathe or drill as a winder, the winding direction will be determined by the machine.
2) If the small windings are at approximately the same potential they can share a layer, otherwise you must provide sufficient insulation between them. Sometimes you may have to separate layers to reduce capacitance between them, or interleave them to reduce leakage inductance, but that is too complex to discuss here.
3) Winding coils side by side is simpler and provides better voltage isolation especially when the sections are different bins on the bobbin. That is essential for high voltage supplies and makes mains voltage safety approval easier.
However it increases the leakage inductance, and that matters for high frequency applications like SMPS. The only reasonable way to know if you can do this is to wind one that way and measure the leakage inductance. If it's out of spec (or even close to it) wind another in layers and compare them.
Any transformer connected to 240V AC has to have sufficient inductance in the primary so that it isn't taking a large magnetization current - just think of the primary and ignore the secondary for now - imagine you are only making an inductor to connect to the AC - you don't want it taking ten amps just by itself.
Of course there is a technical reason for not taking ten amps - this will almost certainly saturate the core and fry.
So, armed with the details you have on the toroid such as the \$A_L\$ value, this will help you understand how many turns are needed to obtain an inductance of (say) 10 henries. 10 henries will have an impedance of about 3000 ohms at 50Hz and will take a current of about 80mA when connected to the 240V AC.
Then you need to decide if this will saturate the core and fry. You Used \$A_L\$ and a target inductance of 10 henries to tell you how many turns you need to wind then, you can calculate the magneto-motive force (ampere-turns). Then, divide MMF by the net length around your toriod to get H (magnetic field strength, ampere-turns per metre) and you are nearly there.
Referring to the BH curve in the toroid's data sheet and using the value of H just calculated determine what magnetic flux density (B) you will be getting from the graphs normally supplied - if it's more than about 0.4 Teslas then you will likely run into saturation problems.
I'm not doing the math but it's going to be a close run thing as to whether this toroid is big enough to tolerate mains voltage across the primary - ferrite toroids are not normally used as regular AC transformers - they are better suited to a much higher frequency (as per what you get in an off-line switcher) because of this issue.
If you don't know what the ferrite is made of forget it (and I mean that).
Winding the secondary is a cinch in comparison but before any advise is given, technical detail of the toroid material is needed.
My EI core size is 32mm*60mm giving 368 W maximum power.I want to make 30 0 30 5A secondary out from transformer. according to my calculation dual winding of 75 turns for 30 0 30 V if i want another 25 0 25 V 5A from same winding with same center tap.Can it possible after dual 62 turn to take wire out for 25 0 25 V and continue up 75 turn which will give 30 0 30 final out from secondary.(Note that I will not use both voltage at same time).
Best Answer
This will work as you suggest but you need to take account of loading effects on the transformer and leakage inductance plays a part in this. The upshot being that you might choose to add a turn or two to all secondary windings or maybe remove a few turns from the primary. This means that open circuit the secondaries might be a little high but, under loading conditions, the output will be nominally correct. It's optional.