The input capacitor has two jobs.
High pass filter: In an audio amplifier you want to capacitor to pass everything above 20hz and block below 20hz. (20hz or what frequency you choose)
Bias: You need the amplifier to work at DC and to work at AC. So at DC the signal source shorts out the input. (not good) At DC the capacitor is OPEN (not there). You need to first build an amplifier that works at DC with all the capacitors open (out of circuit). The two input resistors are chosen so the base is at a happy voltage (with no AC signal involved). When a AC signal is applied it sits on the DC point. If the base is at 2 volts DC, then the input signal, of +/-1volt will, through the capacitor, will appear at the Base as the DC and AC added together. 2V with + and - 1 volts. (+1V to +3V)
Transistors are not as straight forward
Yes and no. The way I am explaining transistors is the simple version. There is a more complex version that we don't need to think of now.
A transistor has two parts.
The diode from B to E is much like any diode. If the B gets 0.6 volts above the E current will flow. If B is less then 0.6V above (or negative) current does not flow.
If there is base to emitter current then the transistor allows C to E current to flow. If this transistor has a gain of 100 (Hfe) then 1mA in the base will allow (cause) 100mA in the C.
In the DC part if you circuit if the input resistors pull the base up to 2 volts the emitter will want to come up to 1.4 volts. (2-0.6). If the E is at ground you will have base current. This will turn on the transistor and it will allow current from C-E that in-turn pull up the E. This current will cause a voltage across the emitter resistor that gives us 1.4V at the E. If too much current flows C to E the E voltage gets too high and causes the B-E voltage to be less than 0.6V and the transistor turns off. If too little current flows the E drops in voltage and the B-E voltage is more than 0.6V and the transistor turns on hard. So there is a balance where the base has 0.6V across it.