The speed of an object is given by:
![v=(d)/(t)](https://img.qammunity.org/2023/formulas/mathematics/college/7bvf02ex7prlyl84jiizv8vikm7s8zddn1.png)
For the first part we know that the helicopter can travel 425 miles against a 935 headwind, the resultant speed of the helicopter will be its speed in still air minus the velocity of the wind, then we have:
![\begin{gathered} v_h-935=(425)/(t) \\ t=(425)/(v_h-935) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/f1w5khxvi1687an0s87luk1cenw8pf7kc1.png)
For the scond part the resultant speed of the helicopter is the velocity of the wind plus the velocity of the helicopter, then we have:
![\begin{gathered} v_h+935=(775)/(t) \\ t=(775)/(v_h+935) \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/4h5v7g02m2yqv68ijyf8i5eff1b6pi0jyw.png)
Since the time is equal we have that:
![\begin{gathered} (425)/(v_h-935)=(775)/(v_h+935) \\ 425(v_h+935)=775(v_h-935) \\ 425v_h+397375=775v_h-724625 \\ 775v_h-425v_h=724625+397375 \\ 350v_h=1122000 \\ v_h=(1122000)/(350) \\ v_h=3205.71 \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/5a6a97yug0ta2d9ezbn3fdrv011hqdqpul.png)
Therefore, the velocity of the helicopter is 3205.72 mph