One engine has a diameter of 2.96 m, the area is 6.88 m2.
Airplane speed of 486 kts is 250 m/s. The density at FL280 is 0.493070 kg/m3.
@ Q1) In case there are 2 engines (not specified), having each an intake flow of (assuming that the column of air is sucked into the engine without inlet spillage (drag)):
$$
\dot m = {\rho\times \dot V = \rho\times A\times v = [kg/m^3]\times[m^2]\times[m/s]=[kg/s]}
$$
$$
\dot m = 0.493070\times6.88\times250=848.1\:[kg/s]
$$
@ Q2) For one engine the thrust is 200 kN (under the assumption that 2 engines are on the aircraft and split the load evenly; from a later comment it became clear that there are 4 engines, the exercise continues for 2 engines; for four engines, the thrust value needs to be halved = 100 kN)
(This is an approximation as we are assuming that the inlet mass flow equals the outlet mass flow, no added fuel, and no pressure difference between the inlet and the outlet.)
$$
T=\dot m \times(v_j - v_0)=[kg\;m/s^2] = [kg/s]\times[m/s] = 200,000 = 848.1\times(v_j-250)
$$
$$
=> v_j = 486\;[m/s]
$$
@ Q3)
$$
\eta_j = \frac{2}{1+\frac{v_j}{v_0}}= \frac{2}{1+\frac{486}{250}}=0.68=68\;\%
$$
I would like to know the correct answers
For four engines you will get 368 m/s (average) jet velocity leading to a 81 % jet efficiency; it is left to the reader to calculate this.
(and what I did wrong).
What went wrong is that the thrust is specified for all engines as a total, you need to work out the thrust per engine. Note that the exercise is a simplified version of reality, assumptions are stated in the answer, but for proper calculation should be considered; however, this would require more input.
