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Answer 1

$\displaystyle \left({3\over7}\right)^{3x+1}=\left({7\over3}\right)^{5x-3}\\\\ \left({3\over7}\right)^{3x+1}=\left({3\over7}\right)^{3-5x}\\\\ 3x+1=3-5x\\ 8x=2\\\\ \mathbf{x={1\over4}}$

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$\displaystyle \log_{\pi}{(x^2+2x+3)}=0\\\\ x^2+2x+3>0\\ D=4-12=-8<0,\;\; x\in\mathbb{R}\\\\ \log_{\pi}{(x^2+2x+3)}=\log_{\pi}{1}\\ x^2+2x+3=1\\ x^2+2x+2=0\\ D=4-8=-4<0,\;\; x_{1},x_{2}\notin\mathbb{R}\\ \sqrt{D}=2i\\\\ x_{1}={-2+2i\over2}=-1+i\\x_{2}={-2+2i\over2}=-1-i\\\\ x_{1},x_{2}\in\mathbb{C}$

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$\displaystyle \log_{3}{(4-2x)}-\log_{3}{2}=2\\\\ 4-2x>0\\ x<2\\\\ \log_{3}{4-2x\over2}=\log_{3}9\\\\ 4-2x=18\\2x=-14\\ \mathbf{x=-7}$