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$2)\; \; \sqrt{25+2\sqrt{84}}=\sqrt{25+2\sqrt{4\cdot 21}} =\sqrt{25+2\cdot 2\cdot \sqrt{21}}=\\\\= \sqrt{4+21+2\cdot 2\cdot \sqrt{21}}= \sqrt{2^2+(\sqrt{21})^2+2\cdot 2\cdot \sqrt{21}} =\\\\=\sqrt{(2+\sqrt{21})^2}=|2+\sqrt{21}|=2+\sqrt{21}$

$1) \sqrt{a\pm \sqrt{b} }= \sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{ \frac{a-\sqrt{a^2-b}}{2}}\\\\ \sqrt{15-\sqrt{29}}=\sqrt{\frac{15+\sqrt{225-29}}{2}}-\sqrt{\frac{15-\sqrt{225-29}}{2}}=\\\\= \sqrt{ \frac{15+14}{2}}-\sqrt{\frac{15-14}{2}}= \sqrt{ \frac{29}{2} } - \sqrt{\frac{1}{2}} = \frac{\sqrt{29}-1}{\sqrt2} = \frac{\sqrt2(\sqrt{29}-1)}{2}$

$ili\; \; \; \sqrt{15-\sqrt{29}}=\sqrt{(a-b)^2}\\\\\left \{ {{a^2+b^2=15} \atop {2ab=\sqrt{29}}} \right. \; \left \{ {{a^2+(\frac{\sqrt{29}}{2a})^2=15} \atop {b=\frac{\sqrt{29}}{2a}}} \right. \; \; \to a=\sqrt{\frac{29}{2}}\; ,\; \; b=\frac{1}{2}$