Will a resistor connected in parallel with a high-pass filter affect the filter cut-off freq?
The "trick" you lack may be using the superposition principle to "ignore" (temporarily) the DC source and consider it to be 0 V while studying only the AC source. Using the superposition principle, you can show that V_out consists of a DC component equal to approx. 1V plus an AC component which depends on Vin and the high-pass filter:

Without \$R_1\$ and the DC source, the cutoff frequency is given by :
$$
\omega_c = \frac{1}{2\pi R_2 C_1} = 0.15 Hz
$$
(This is a typical RC circuit)

Adding \$R_1\$ and the DC source, the circuit changes a bit, since you are adding a resistor parallel to \$R_2\$, the cutoff frequency is now
$$
\omega_c = \frac{1}{2\pi (R_2//R_1) C_1} = 0.238 Hz
$$
Since \$R_1//R_2\$ is lower than \$R_2\$, you increased the cutoff frequency, but it's still two decades lower than your expected 50 Hz so it shouldn't be an issue.