If $x^{x^4}=4$, then what is the value of $x^{x^2}+x^{x^8}$?

Just note that $x^{x^4}$ is increasing for $x>1$ and is $<1$ for $x<1$. The unique positive solution to $x^{x^4}=4$ is $x=\sqrt2$.

$x^{x^{4}}=4$. Let us replace 4 of the exponent of left hand side by $x^{x^{4}}$ then equation becomes $x^{x^{x^{x^{4}}}}=4$. Repeating this process infinitely we get $x^{x^{x^{x^{x..}}}}=4$. Now We replace the exponent of left hand side by 4 and equation now become $x^{4}=4$ hence $x=\sqrt{2}$ or $x=-\sqrt{2}$ . So $x^{x^{2}}+x^{x^{8}} =258$