1. for auto baudrate detection you can use a simple poling method. The transmitter - PC, sends a known character for example <LF> - hex code is 0x0a to 8052. You set the baud rate to 300 baud, get one character, if not <LF>, increase baud rate and try again. When you get <LF> from the PC, you acknowledge with the character <ACK>, or similar.

2. digital circuits have a push-pull output consisting of a P-channel and N-channel MOS transistor. The gates are connected together. When output is high, output stage driving voltage is low, so only the P-channel transistor (at the VCC) is conducting, so output voltage is tied to VCC.
When output is low, output stage driving voltage is high, so only the N-channel transistor (at the GND) is conducting, so output voltage is tied to GND.
In both cases, there is no current from the VCC to GND.
But consider the situation, when the driving voltage is in the middle - one half of VCC, in this case, both transistors are conducting for a while. This makes a current spike from VCC to GND, for every switch from 0-->1 or 1-->0, you get one current spike. Now it depends on the frequency of the output changes, how many current spikes you have and therefore, what is the power consumption of a digital circuit.