[tex]\displaystyle\bf\\2)\\\\(2-3i)-\left(\frac{1}{3}+\frac{2}{5}i\right)=\\\\\\=\left(2-\frac{1}{3}\right)+\left(-3i-\frac{2}{5}i\right)=\\\\\\=\left(2-\frac{1}{3}\right)+\left(-3i-\frac{2}{5}i\right)=\\\\\\=\left(2-\frac{1}{3}\right)-\left(3+\frac{2}{5}\right)i=\\\\\\=\frac{2\times3-1}{3}-\frac{3\times5+2}{5}i=\\\\\\=\boxed{\bf\frac{5}{3}-\frac{17}{5}i}[/tex]
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[tex]\displaystyle\bf\\3)\\\\\frac{2-3i}{7+i}=\\\\\\=\frac{(2-3i)(7-i)}{(7+i)(7-i)}=\\\\\\=\frac{14-21i-2i+3i^2}{7^2-i^2}=\\\\\\=\frac{14-23i-3}{49+1}=\\\\\\=\frac{14-3-23i}{50}=\\\\\\=\frac{11-23i}{50}=\\\\\\=\boxed{\bf\frac{11}{50}-\frac{23}{50}i}[/tex]
