The total active power of the system is given by the sum of the active powers of the induction motor and synchronous condenser:

P = P1 + P2

where P1 is the active power of the induction motor and P2 is the active power of the synchronous condenser.

P1 is given by:

P1 = S1 × PF1

where S1 is the apparent power of the induction motor and PF1 is the power factor of the induction motor.

Substituting the given values, we get:

P1 = 1000 × 0.866 = 866 kW

Similarly, P2 is given by:

P2 = S2 × PF2

where S2 is the apparent power of the synchronous condenser and PF2 is the power factor of the synchronous condenser.

Substituting the given values, we get:

P2 = 750 × 0.707 = 530.25 kW

Therefore, the total active power of the system is:

P = 866 + 530.25 = 1396.25 kW

The total reactive power of the system is given by the sum of the reactive powers of the induction motor and synchronous condenser:

Q = Q1 + Q2

where Q1 is the reactive power of the induction motor and Q2 is the reactive power of the synchronous condenser.

Q1 is given by:

Q1 = P1 × tan(cos?¹PF1)

Substituting the given values, we get:

cos?¹PF1 = cos?¹0.866 = 30°

tan30° = 0.577

Q1 = 866 × 0.577 = 500 kVAR

Q2 is given by:

Q2 = P2 × tan(cos?¹PF2)

Substituting the given values, we get:

cos?¹PF2 = cos?¹0.707 = 45°

tan45° = 1

Q2 = 750 × 1 = 750 kVAR (since PF2 is leading, Q2 is negative)

Therefore, the total reactive power of the system is:

Q = 500 - 750 = -250 kVAR (since Q2 is negative)

The power factor of the total load supplied by the mains is given by:

cos? = P / √(P² + Q²)

Substituting the calculated values, we get:

cos? = 1396.25 / √(1396.25² + (-250)²) = 0.99 leading