Output waveform of this circuit
This question is about 1 month old, why did it pop to the top of the queue today?  

In any case, I will attempt to answer

1) with the capacitor in place, the 2Vpp sine wave is impressed upon the 2.5V DC offset bias point by the voltage divider made up of the two resistors. However, the C and R make a first-order high pass filter, where C = 1 \$\mu\$F and R is the parallel combination of 100//100 = 50 \$\Omega\$ = Rth
(because that voltage divider has a Thevinen equivalent of 50\$\Omega\$)

$$ F_c = {{1} \over {2\pi\cdot R\cdot C }} = {1 \over {2\pi\cdot 50\cdot 1e^{-6}}} = 3.183\space \mathrm{KHz}$$

You said your input signal can be "of any frequency \$f\$". What this means is that the value seen at the point 'B' will be attenuated at any frequency below the cut-off frequency of 3.183 KHz, and the attenuation will be 20db per decade below \$F_c\$. This attenuated value will then be offset at point 'B' to 2.5V by the voltage divider. 

In order to determine mathematically what the voltage will be at point 'B' at any frequency, we need to consider that the Cin and the Rth form yet another voltage divider, where the Cin has a frequency dependent resistance, with units in ohms and called reactance, which is equal to the following function of \$f\$ : 

$$ Xc(f) = {{1} \over {2\pi\cdot f\cdot C}}$$



Since the voltage divider equation is \$ V_{out} = { V_{in} \cdot \left ( R_{bot}\over {R_{top} + R_{bot}} \right ) } \$ , with \$R_{bot}\$ = Rth and \$R_{top} = Xc(f)\$, \$V_{in} = A\cdot \sin(\omega t) \$, A = \$1\$ Volt peak, \$\omega\$ is the chosen frequency in radians/second, called the angular frequency, \$\omega = 2\cdot\pi f\$ and \$t\$ is time. 

With all this, we can put it all together to come up with a transfer function from the input sine to B at any given frequency over time:
$$
B(f,t) = {V_{peak}\sin( 2\pi f t)} \cdot \left ( {50}\over {({2\pi\cdot f\cdot 1e^{-6}}) + 50} \right ) +2.5
$$

Finally, we need to consider the source impedance.  Since you didn't say, I am assuming that the voltage source is ideal and has 0 impedance.

2) with a DC voltage of 1V at the capacitor, then nothing happens at point B, it remains at 2.5V, and the capacitor is charged to 1.5V.  That's the DC operating point.  Now, if you want to know what happens at time t=0, when you first apply the 1V DC, and time after that, then there will appear a discharge curve at point B from 3.5V down to 2.5V, exponentially, opposite to the generic capacitor charge equation : 

$$
V_c(t) = V_0\cdot e^{-t\over{rc}}
$$

This is because when 1V is first applied, C is discharged and appears as a short circuit, so the 1V is added to the 2.5V offset at 'B' to make 'B' 3.5V. As the capacitor charges up according to the charging equation, the voltage across the capacitor subtracts from the total voltage at B, until the capacitor reaches 1.5V, and 'B' is at 2.5V again.   In other words, the curve discharges down from 3.5V to 2.5V exponentially. After that it just stays at 2.5V, the DC operating point.