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$1.\;\left[11+\frac32\cdot\left(\frac15-\frac13\right)\right]:5^{-1}\\1)\;\frac15-\frac13=\frac{3-5}{15}=-\frac2{15}\\2)\;\frac32\cdot\left(-\frac2{15}\right)=-\frac15\\3)\;11+\left(-\frac15\right)=11-\frac15=10\frac45=\frac{54}5\\4)\;\frac{54}5:5^{-1}=\frac{54}5:\frac15=\frac{54}5\cdot5=54$

$2.\;0,25\cdot2^4\cdot5-27^{\frac13}=\frac14\cdot16\cdot5-\sqrt[3]{27}=4\cdot5-3=20-3=17\\\\3.\;2^6\cdot2^{\frac12}\cdot2^{\frac14}\cdot2^{\frac16}=2^{6+\frac12+\frac14+\frac16}=2^{\frac{72+6+3+1}{12}}=2^{\frac{83}{12}}=2^{6\frac{11}{12}}=\\=2^{6\frac{11}{12}+\frac1{12}-\frac1{12}}=2^{7-\frac1{12}}=\frac{2^7}{2^{\frac12}}=\frac{128}{\sqrt2}=\frac{128\sqrt2}{\sqrt2\sqrt2}=\frac{128\sqrt2}{2}=64\sqrt2$

$4.\;\left[\frac67\left(\frac12-\frac13\right)+1\frac67\right]^{-2}\\1)\;\frac12-\frac13=\frac{3-2}6=\frac16\\2)\;\frac67\cdot\frac16=\frac17\\\frac17+1\frac67=2\\2^{-2}=\frac1{2^2}=\frac14\\\\5.\;\frac{2^{\frac23}}{4^{\frac56}}=\frac{2^{\frac23}}{(2^2)^{\frac56}}=\frac{2^{\frac23}}{2^{\frac53}}=2^{\frac23-\frac53}=2^{-1}=\frac12\\\\6.\;\left(b^{-1}\cdot b^9\right)^{-2}\cdot b^{-3}=(b^{-1+9})^{-2}\cdot b^{-3}=(b^8)^{-2}\cdot b^{-3}=b^{-16}\cdotb^{-3}=b^{-19}$

$7.\;a^{\frac23}\cdot a^{-\frac54}=a^{\frac23-\frac54}=a^{-\frac7{12}}$