a/ \(2^3+3\cdot\left(\dfrac{2}{3}\right)^0-2^{-2}+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8=8+3\cdot1-\dfrac{1}{2^2}+4\cdot2-8=\left(8-8\right)+3+8-\dfrac{1}{4}=11-\dfrac{1}{4}=\dfrac{43}{4}\)
b/ \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2=1:\dfrac{1}{3}-1+\dfrac{1}{4}\cdot\dfrac{1}{2}=3-1+\dfrac{1}{8}=2+\dfrac{1}{8}=\dfrac{17}{8}\)