[tex]\displaystyle\bf\\5)\\a)\\10\cdot\left\{18^2:324+2\cdot\left[\Big(2^2\cdot3\Big)^{15}:\Big(2^{29}\cdot3^{15}\Big)+1^24\right]\right\}=\\\\\\=10\cdot\left\{324:324+2\cdot\left[\Big(2^{2\times15}\cdot3^{15}\Big):\Big(2^{29}\cdot3^{15}\Big)+1\right]\right\}=\\\\\\=10\cdot\left\{1+2\cdot\left[\Big(2^{30}\cdot3^{15}\Big):\Big(2^{29}\cdot3^{15}\Big)+1\right]\right\}=[/tex]
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[tex]\displaystyle\bf\\=10\cdot\left\{1+2\cdot\left[\Big(2^1\cdot3^0\Big)+1\right]\right\}=\\\\\\=10\cdot\left\{1+2\Big[2\cdot1+1\Big]\right\}=\\\\\\=10\cdot\left\{1+2\Big[2+1\Big]\right\}=\\\\\\=10\cdot\Big\{1+2\cdot3\Big\}=\\\\\\=10\cdot\Big\{1+6\Big\}=\\\\\\=10\cdot7=\boxed{\bf70}[/tex]
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[tex]\displaystyle\bf\\b)\\\left[2^{12}\cdot2^{18}+5^{70}:5^{10}-\Big(3^{25}\Big)^{\b2}\right]:\left[4^2\cdot2^3\cdot2^{23}+\Big(5^{15}\Big)^{\b2^{\b2}}-\Big(3^2\Big)^{\b5^{\b2}}\right]=\\\\\\=\left[2^{12+18}+5^{70-10}-\Big(3^{25\times2}\Big)\right]:\left[\Big(2^2\Big)^2\cdot2^3\cdot2^{23}+\Big(5^{15}\Big)^4-\Big(3^2\Big)^{25}\right]=\\\\\\=\Big[2^{30}+5^{60}-3^{50}\Big]:\Big[2^{2\times2}\cdot2^3\cdot2^{23}+5^{15\times4}-3^{2\times25}\Big]=[/tex]
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[tex]\displaystyle\bf\\=\Big[2^{30}+5^{60}-3^{50}\Big]:\Big[2^{4+3+23}+5^{60}-3^{50}\Big]=\\\\\\=\Big[2^{30}+5^{60}-3^{50}\Big]:\Big[2^{30}+5^{60}-3^{50}\Big]=\boxed{\bf1}[/tex]
