Solution:

Consider ‘r’ to be a positive real number and let ‘a’ be a fixed real number. Then, we can write:

|x + a| > r ⟺ x > r – a or x < – (a + r)

Here, a = 1/3 and r = 8/3

x > 8/3 – 1/3 or x < – (8/3 + 1/3)

x > (8-1)/3 or x < – (8+1)/3

x > 7/3 or x < – 9/3

Therefore, x > 7/3 or x < – 3

x ∈ (7/3, ∞) or x ∈ (–∞, -3)

∴ x ∈ (–∞, -3) ∪ (7/3, ∞)