a)

${\log }_{\frac{1}{2}}8=x\Rightarrow {\left(\frac{1}{2}\right)}^x=8$

${\left(\frac{1}{2}\right)}^x=8$

$2^{-x}=2^3$

$-x=3$

$x=\left(-3\right)$

b)

${\log }_{\sqrt{2}}4=x\Rightarrow {\sqrt{2}}^x=4$

${\sqrt{2}}^x=4$

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

$2^{\frac{1}{2}x}=2^2$

$\frac{1}{2}x=2$