${27}^{\frac{1}{3}}$

= $\sqrt[3]{27}$

= $3$

$3^{38}$ : $3^{35}$

= $3^{38-35}$

= $3^3$

= 27

${\mathrm{0,008}}^{\frac{1}{3}}$

= $\sqrt[3]{\mathrm{0,008}}$

= $\mathrm{0,2}$

${\left(4^3\right)}^{\frac{1}{2}}$

= $4^{3·\frac{1}{2}}$

= $4^{\frac{1}{2}·3}$

= ${\left(4^{\frac{1}{2}}\right)}^3$

= ${\left(\sqrt{4}\right)}^3$

= $2^3$

= $8$

Hier ist es eben wichtig, dass man die Potenzgesetze erkennt und dann rückwärts anwendet:

${(\sqrt[4]{x}·\sqrt[12]{x^{15}})}^6$

= ${(x^{\frac{1}{4}}\cdot x^{\frac{15}{12}})}^6$

= ${(x^{\frac{1}{4}}\cdot x^{\frac{5}{4}})}^6$

= ${\left(x^{\frac{1}{4}+\frac{5}{4}}\right)}^6$

= ${\left(x^{\frac{3}{2}}\right)}^6$

= $x^{\frac{3}{2}·6}$

= $x^9$