$\begin{array}{l}1)\\a)\ \dfrac{18}{37}+\dfrac8{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac23\\=\dfrac{18}{37}+\dfrac13+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac23\\=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac13+\dfrac23\right)-1\dfrac{23}{24}\\=1+1-1\dfrac{23}{24}\\=1\dfrac{24}{24}-1\dfrac{23}{24}\\=\dfrac1{24}\\\,\\b)\ \left(3\dfrac13+2,5\right)\div\left(3\dfrac16-4\dfrac15\right)-\dfrac{11}{31}\\=\left(\dfrac43+\dfrac52\right)\div\left(\dfrac{19}6-\dfrac{21}5\right)-\dfrac{11}{31}\\=\dfrac{23}6\div\dfrac{-31}{30}-\dfrac{11}{31}\\=\dfrac{-115}{31}-\dfrac{11}{31}\\=\dfrac{-126}{31}\\\,\\c)\ \left[6+\left(\dfrac12\right)^3-\left|-\dfrac12\right|\right]\div\dfrac3{12}\\=\left(6+\dfrac18-\dfrac12\right)\div\dfrac14\\=\left(\dfrac{48}8+\dfrac18-\dfrac48\right)\cdot4\\=\dfrac{45}{48}\cdot4\\=\dfrac{45}2\\\,\\d)\ \dfrac25\cdot\dfrac13-\dfrac2{15}\div\dfrac15+\dfrac35\cdot\dfrac13\\=\dfrac2{15}-\dfrac23+\dfrac15\\=\dfrac2{15}-\dfrac{10}{15}+\dfrac3{15}\\=\dfrac{-1}3\\\,\\e)\ \dfrac{-4}9\cdot\left(\dfrac56\cdot90\%-8\div0,32\right)\\=\dfrac{-4}9\cdot\left(\dfrac56\cdot\dfrac9{10}-8\div\dfrac{8}{25}\right)\\=\dfrac{-4}9\cdot\left(\dfrac34-25\right)\\=\dfrac{-4}9\cdot\dfrac{-97}4\\=\dfrac{97}9\\\,\\f)\ \left(\dfrac25\right)^2+5\dfrac12\cdot(4,5-2)+\dfrac{2^3}{-4}\\=\dfrac{4}{25}+\dfrac{11}2\cdot2,5+\dfrac8{-4}\\=\dfrac4{25}+\dfrac{11}2\cdot\dfrac52-2\\=\dfrac{4}{25}+\dfrac{55}4-2\\=\dfrac{16}{100}+\dfrac{1375}{100}-\dfrac{200}{100}\\=\dfrac{1191}{100}\\\,\\2)\\\quad A=49\dfrac8{23}-\left(5\dfrac7{32}+14\dfrac8{23}\right)\\\to A=49\dfrac8{23}-5\dfrac7{23}-14\dfrac8{23}\\\to A=\left(49\dfrac8{23}-14\dfrac8{23}\right)-5\dfrac7{32}\\\to A=35-5\dfrac7{23}\\\to A=34\dfrac{23}{23}-5\dfrac7{23}\\\to A=29\dfrac{16}{23}\\\,\\\quad B=71\dfrac{38}{45}-\left(43\dfrac8{45}-1\dfrac{17}{57}\right)\\\to B=71\dfrac{38}{45}-43\dfrac8{45}+1\dfrac{17}{57}\\\to B=28\dfrac23+1\dfrac{17}{57}\\\to B=28\dfrac{38}{57}+1\dfrac{17}{57}\\\to B=29\dfrac{55}{57}\\\,\\\quad C=\dfrac{-3}7\cdot\dfrac59+\dfrac49\cdot\dfrac{-3}7+2\dfrac37\\\to C=\dfrac{-3}7\cdot\left(\dfrac59+\dfrac49\right)+2\dfrac37\\\to C=\dfrac{-3}7\cdot1+2\dfrac37\\\to C=\dfrac{-3}7+2\dfrac37\\\to C=2\\\,\\\quad D=\left(19\dfrac58\div\dfrac7{12}-13\dfrac14\div\dfrac7{12}\right)\cdot\dfrac45\\\to D=\left(19\dfrac58\cdot\dfrac{12}7-13\dfrac48\cdot\dfrac{12}7\right)\cdot\dfrac45\\\to D=\dfrac{12}7\cdot\left(19\dfrac58-13\dfrac48\right)\cdot\dfrac45\\\to D=\dfrac{12}7\cdot6\dfrac18\cdot\dfrac45\\\to D=\dfrac{12}7\cdot\dfrac{49}8\cdot\dfrac45\\\to D=\dfrac{42}5\\\,\\\quad E=\left(1+\dfrac12\right)\cdot\left(1+\dfrac13\right)\cdot\left(1+\dfrac14\right)\cdots\left(1+\dfrac1{99}\right)\\\to E=\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\\\to E=\dfrac{3.4.5\ldots100}{2.3.4\ldots99}\\\to E=\dfrac{100.(3.4.5\ldots99)}{2.(3.4.5\ldots99)}\\\to E=\dfrac{100}2\\\to E=50 \end{array}$
