\begin{enumerate}
\item $\dfrac{1+\i}{1-\i}=\dfrac{(1+\i)(1+\i)}{(1-\i)(1+\i)}=\dfrac{2\i}{2}=\boxed{\i}$.
\item $\dfrac{1}{3+5\i}=\dfrac{1(3-5\i)}{(3+5\i)(3-5\i)}=\dfrac{3-5\i}{3^2+5^2}=\boxed{\dfrac{3}{34}-\dfrac{5}{34}\i}$
\item $\dfrac{2+\i}{3-4\i}=\dfrac{(2+\i)(3+4\i)}{(3-4\i)(3+4\i)}=\dfrac{2+11\i}{3^2+4^2}=\boxed{\dfrac{2}{25}+\dfrac{11}{25}\i}$
\item $\dfrac{\i-1}{3+2\i}=\dfrac{(-1+\i)(3-2\i)}{(3+2\i)(3-2\i)}=\dfrac{-1+5\i}{3^2+2^2}=\boxed{-\dfrac{1}{13}+\dfrac{5}{13}\i}$
\item $\dfrac{5+3\i}{2\i-1}=\dfrac{(5+3\i)(-1-2\i)}{(-1+2\i)(-1-2\i)}=\dfrac{1-13\i}{(-1)^2+2^2}=\boxed{\dfrac{1}{5}-\dfrac{13}{5}\i}$
\item $\dfrac{3\i-4}{4\i-3}=\dfrac{(-4+3\i)(-3-4\i)}{(-3+4\i)(-3-4\i)}=\dfrac{24+7\i}{(-3)^2+4^2}=\boxed{\dfrac{24}{25}+\dfrac{7}{25}\i}$
\end{enumerate}

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