Trigonometric Identity Relation Involving Fourth Powers of Sine and Cosine

To determine the correct options, we can apply Cauchy-Schwarz inequality in Engel form (Titu's Lemma) or solve the equation directly.

Method 1: Using Cauchy-Schwarz Inequality

By the Cauchy-Schwarz inequality (or Titu's Lemma), for any real numbers:
$\frac{(\sin^2 x)^2}{2} + \frac{(\cos^2 x)^2}{3} \ge \frac{(\sin^2 x + \cos^2 x)^2}{2 + 3}$

Since $\sin^2 x + \cos^2 x = 1$, the right-hand side becomes:
$\frac{1^2}{5} = \frac{1}{5}$

We are given:
$\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$

Since the equality condition holds in the inequality, the terms must be proportional:
$\frac{\sin^2 x}{2} = \frac{\cos^2 x}{3}$

From this ratio, we get:
$\frac{\sin^2 x}{\cos^2 x} = \frac{2}{3} \implies \tan^2 x = \frac{2}{3}$
Thus, Option (A) is correct, and Option (C) is incorrect.

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Method 2: Finding $\sin^2 x$ and $\cos^2 x$

From the proportionality relation:
$\frac{\sin^2 x}{2} = \frac{\cos^2 x}{3} = \frac{\sin^2 x + \cos^2 x}{2 + 3} = \frac{1}{5}$

This yields:
$\sin^2 x = \frac{2}{5}$
$\cos^2 x = \frac{3}{5}$

Now, we evaluate the expression in options (B) and (D):
$\frac{\sin^8 x}{8} + \frac{\cos^8 x}{27} = \frac{(\sin^2 x)^4}{8} + \frac{(\cos^2 x)^4}{27}$

Substitute the values of $\sin^2 x$ and $\cos^2 x$:
$\frac{\sin^8 x}{8} + \frac{\cos^8 x}{27} = \frac{\left(\frac{2}{5}\right)^4}{8} + \frac{\left(\frac{3}{5}\right)^4}{27}$
$= \frac{\frac{16}{625}}{8} + \frac{\frac{81}{625}}{27}$
$= \frac{2}{625} + \frac{3}{625} = \frac{5}{625} = \frac{1}{125}$

Thus, Option (B) is correct, and Option (D) is incorrect.

Conclusion:

The correct options are A and B.