If sin⁴x/2 + cos⁴x/3 = 1/5, then

sin⁴x/2 + cos⁴x/3 = 1/5
⇒sin⁴x/2 + (1 - sin²(x))²/3 = 1/5
⇒sin⁴x/2 + 1 + sin⁴x - 2sin²(x)/3 = 1/5
⇒3sin⁴x + 2 + 2sin⁴x - 4sin²(x)/6 = 1/5
⇒5sin⁴x - 4sin²(x) + 2/6 = 1/5
⇒25sin⁴x - 20sin²(x) + 10 = 6
⇒25sin⁴x - 20sin²(x) + 4 = 0
Let sin²x = t
⇒25t² - 20t + 4 = 0
⇒(5t - 2)² = 0
⇒5t - 2 = 0
⇒t = 2/5
⇒sin²x = 2/5, cos²x = 3/5, tan²x = 2/3
Hence,
sin⁸x/8 + cos⁸x/27 = (2/5)⁴/8 + (3/5)⁴/27 = 16/25⁴ + 27/25⁴ = 1125