Answer:
<em>a) 0.67 rad/sec in the clockwise direction.</em>
<em>b) 98.8% of the kinetic energy is lost.</em>
Explanation:
Let us take clockwise angular speed as +ve
For first cylinder
rotational inertia 
= 2.0 kg-m^2
angular speed ω = +5.0 rad/s
For second cylinder
rotational inertia = 1.0 kg-m^2
angular speed = -8.0 rad/s
The rotational momentum of a rotating body is given as = ω
where is the rotational inertia
ω is the angular speed
The rotational momenta of the cylinders are:
for first cylinder = ω = 2.0 x 5.0 = 10 kg-m^2 rad/s
for second cylinder = ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s
The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = <em>2 kg-m^2 rad/s</em>
When they are coupled coupled together, their total rotational inertia 
= 1.0 + 2.0 = 3 kg-m^2
Their final angular rotational momentum after coupling = 
where is their total rotational inertia
= their final angular speed together
Final angular momentum = 3 x = 3
According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum
this means that
2 = 3
= final total angular speed of the coupled cylinders = 2/3 = <em>0.67 rad/s</em>
From the first statement, <em>the direction is clockwise</em>
b) Rotational kinetic energy = 
where is the rotational inertia
is the angular speed
The kinetic energy of the cylinders are:
for first cylinder = = 
= 25 J
for second cylinder = 
= 32 J
Total initial energy of the system = 25 + 32 = 57 J
The final kinetic energy of the cylinders after coupling = 
where
where is the total rotational inertia of the cylinders
is final total angular speed of the coupled cylinders
Final kinetic energy = 
= 0.67 J
kinetic energy lost = 57 - 0.67 = 56.33 J
percentage = 56.33/57 x 100% = <em>98.8%</em>
