**Given:**
Initial angular velocity (ω₀) = 2 rad/s
Angular acceleration (α) = 3t² - 3

**To Find:**
Angular velocity (ω) and angular displacement (θ) when time (t) is 5s.

**Solution:**

1. First, let's find the angular velocity (ω) at time t = 5s.

We know that the angular acceleration (α) is the derivative of angular velocity (ω) with respect to time (t).

α = dω/dt

Given α = 3t² - 3

Integrating both sides with respect to t, we get:

∫α dt = ∫(3t² - 3) dt

ω = ∫(3t² - 3) dt

ω = t³ - 3t + C

where C is the constant of integration.

2. Now, we can find the value of C using the initial angular velocity (ω₀) at t = 0.

ω₀ = 2 rad/s

Substituting t = 0 and ω = 2 into the equation ω = t³ - 3t + C, we get:

2 = (0)³ - 3(0) + C

2 = C

Therefore, the equation for angular velocity (ω) becomes:

ω = t³ - 3t + 2

3. Next, let's find the angular displacement (θ) at time t = 5s.

The angular displacement (θ) can be obtained by integrating the angular velocity (ω) with respect to time (t).

θ = ∫ω dt

θ = ∫(t³ - 3t + 2) dt

θ = (1/4)t⁴ - (3/2)t² + 2t + D

where D is the constant of integration.

4. To find the value of D, we can use the initial condition where t = 0 and θ = 0.

θ = (1/4)(0)⁴ - (3/2)(0)² + 2(0) + D

0 = 0 - 0 + 0 + D

D = 0

Therefore, the equation for angular displacement (θ) becomes:

θ = (1/4)t⁴ - (3/2)t² + 2t

5. Finally, we can find the angular velocity (ω) and angular displacement (θ) at time t = 5s by substituting t = 5 into the respective equations:

ω = (5)³ - 3(5) + 2
= 125 - 15 + 2
= 112 rad/s

θ = (1/4)(5)⁴ - (3/2)(5)² + 2(5)
= (1/4)(625) - (3/2)(25) + 10
= 156.25 - 37.5 + 10
= 128.75 radians

**Answer:**
The angular velocity at t = 5s is 112 rad/s and the angular displacement at t = 5s is 128.75 radians.