Purpose of 2 resistors in high pass filter?
This is basically a voltage divider between \$R_2\$ and \$R_1, C_1\$ in series, so the transfer function is the following: $${V_{out} \over V_{in}}={R_2 \over{R_2+R_1+{1\over j\omega C_1}}}={j\omega R_2C_1 \over 1+j\omega C_1(R_1+R_2)}$$

The first resistor limits the current into the capacitor, therefore it can charge slower, with a time constant of \$\tau=(R_1+R_2)C_1\$.

The cutoff frequency is smaller than without \$R_1\$: instead of $$f_c={1\over 2\pi R_2C_1}$$ it is $$f_c={1\over 2\pi(R_1+R_2)C_1}$$