$\text{a)}\sqrt{24}+3\sqrt{48}-2\sqrt{75}=\sqrt{4\cdot 6}+3\sqrt{16\cdot 3}-2\sqrt{25\cdot 3}=2\sqrt{6}+12\sqrt{3}-10\sqrt{3}=2\sqrt{6}+2\sqrt{3}=2\left(\sqrt{6}+\sqrt{3}\right)$  

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$\text{b)}{\left(2\sqrt{2}-3\sqrt{10}\right)}^2={\left(2\sqrt{2}\right)}^2-2\cdot 2\sqrt{2}\cdot 3\sqrt{10}+{\left(3\sqrt{10}\right)}^2=8-12\sqrt{20}+90=8-24\sqrt{5}+90=98-24\sqrt{5}$  

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$\text{c)}\sqrt[4]{\frac{1}{16}}\sqrt[3]{-\frac{1}{8}}=\sqrt[4]{{\left(\frac{1}{2}\right)}^4}\cdot \left(-\frac{1}{2}\right)=\frac{1}{2}\cdot \left(-\frac{1}{2}\right)=-\frac{1}{4}$  

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$\text{d)}{\left(\sqrt[4]{\frac{1}{16}}\right)}^2-\sqrt[3]{-\frac{1}{8}}={\left(\sqrt[4]{{\left(\frac{1}{2}\right)}^4}\right)}^2-\left(-\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^2-\left(-\frac{1}{2}\right)=\frac{1}{4}+\frac{1}{2}=\frac{1}{4}+\frac{2}{4}=\frac{3}{4}$