We have,
$$
\begin{aligned}
& z_1=1-2 i, z_2=1+i, \\
& z_3=3+4 i
\end{aligned}
$$
Now, $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}=\left(\frac{1}{1-2 i}+\frac{3}{1+i}\right)\left(\frac{3+4 i}{1+i}\right)$
$$
\begin{aligned}
& =\frac{(1+i)+3(1-2 i)}{(1-2 i)(1+i)} \times \frac{3+4 i}{1+i}=\frac{4-5 i}{3-i} \times \frac{3+4 i}{1+i} \\
& =\frac{12+16 i-15 i+20}{3+3 i-i+1}=\frac{32+i}{4+2 i} \\
& =\frac{(32+i)(2-i)}{2(2+i)(2-i)}=\frac{64+2 i-32 i+1}{2(4+1)} \\
& =\frac{65-30 i}{10}=\frac{13}{2}-3 i
\end{aligned}
$$