How to choose V_out resistors for LT8494 SEPIC
You don't say what output voltage you require.

EDIT - with a range of 1.2 - 12V you don't need the multiplier, you can use the circuit at the top of page 19 of the data sheet.  Change the feedback resistors (316k) to match your voltage needs.

This circuit goes beyond a simple SEPIC converter which wouldn't have D1-D4, D7 and D8 etc. Normally the output would be from C2. If you don't need as much as that you may be able to get away with fewer diodes or no multiplier at all.

Those other components have been added to increase the output voltage up to 60V. They form a DC multiplier (sometimes called a Cockcroft-Walton multiplier).

One common reason for using the SEPIC configuration is that it is tolerant of a short-circuit on the output - the normal boost converter has a direct path from Vin to the output so a short will probably destroy the inductor.  In the SEPIC configuration the signal passes though the capacitor C3 so no DC can flow. The inductor L2 provides a DC path for the output current to flow through the diode D5 while allowing the AC voltage from C3 to couple to D5.

Another advantage of the SEPIC is that the output can be higher or lower than the input voltage, a normal boost circuit can only produce voltages higher than the input and a buck only produce voltages less than the input.

R1, R2 and R3 are to provide the feedback signal to the FB pin of the IC.  The value needs to be low enough so that it is not affected by the input current of the IC (only 20nA from the data sheet) but high enough so the power consumed by the resistors is not significant relative to the load current.

The FB pin will be at 1.25V when the circuit is in control of the output voltage, if the voltage goes higher than that it will try to reduce the output, if it goes lower it will increase the output.  The DAC input to the resistor network will modify the voltage at the output to achieve that balance. Just treat it as a resistor network to determine the range of control that you need.  Increasing the voltage from the DAC will cause the output voltage to reduce. At the specified output voltage the output from the resistor network should equal 1.25V.

It looks like the BIAS input pin is used to power the internal circuitry for situations where the input - it is explained on page 14 of the data sheet.  This is specific to this device.