I Need 5Vac From Distribution Transformer Line.

Please do not use abbreviations (unless you define them in your post). It makes it difficult to understand - what does w/c stand for? It usually means "water closet", familiar to architects and sewage engineers and also non-engineers? Do you mean "which"? Was it worth saving a second typing only to cause a communication failure? What is MCU? Maybe it means "Micro-Controller Unit" to some used to the jargon of silicon computers, but it might mean Main Control Unit or Mains Controlled Unit or Maximum Cash Utilisation in another area. As a young engineer, it was impressed on me (time after time!) that, when sending a message, you must take care it cannot be understood two ways or use jargon clear to you but not to the man at the other end.

To answer your questions (you did have the sense to send a diagram) :-

1. In a distribution transformer which is overloaded, the current will exceed rated value for continuous use, but the voltage will not fall much from normal. In addition, most of the variable losses of the transformer depend on the load current squared [ Often called I²R losses]. Since overheating is the final limit on overload, the voltage fall is less reliable - it might just be a fall in primary supply.
2. The CT output current is not small. The output current of a distribution transformer is usually 50 amps or more at 240V. Standard CTs have 5 amp secondaries at rated continuous primary current - they are usually described as 100/5A or 1000/ 5 amp to make this clear, writing 20:1 ratio or 200:1 ratio gives no idea of rated values. CTs with 1 amp secondary current are sometimes used when [secondary] cable volt drop is a problem (but suffer from the much higher voltages during overcurrent).
3. Your question 1 -- The connection and use of a fixed resistor to get a proportional voltage is valid and often used. But if you want to power a small computer from the voltage you need direct current -which requires rectifiers, reservoir capacitors and a voltage regulator. Also, a short circuit of the transformer secondary (at a load or line) would cause typically up to 25 times rated current through the CT - and consequently 25 times your 5 Volts = 125 volts. Also, a short can last as long as 3 seconds before it is removed by protection. Note that 125 volts x 5 amps = 625 watts and the "5V drop resistor" has to absorb this power for 3 seconds without damage or affect on its ohmic value. Although saturation of the CT iron core may limit the rms (root mean square) current to less than 25 times, great voltages can be produced as the core comes out of saturation. Another problem is that to get 5 volts at 5amps you need a 5 x 5 = 25 volt-amp [VA] rated CT, which is far more power than you need for the MCU.
4. Your question 2 -- A modern microcontroller MCU can use very little power - the error is probably negligible compared to the CT error which starts at 5% and gets expensive beyond 1%. The reason is that CTs usually have iron cores, which require a magnetizing current in the primary, which never reaches the secondary.
5. Your question 3 -- Main CTs for power systems usually have just one secondary, two wires. They can have more windings in principle, but there is extra error due to balance between the windings. Measuring circuits, fed from the primary CT, do use more than 2 windings, when it is useful.
6. Your final sentence seems to be a question, but has no question mark (?). Always remember that someone whose native language is not English may understand ? better than words! What am I to suggest? Show a figure of what (what does "it" stand for here?)?