The descending order of the moduli of

$$\begin{gathered} z_1=(3-4i)(4+3i) \\ z_2=\frac{3+4i}{1+i} \\ {z}_3=\frac{\left(3+i\right)\left(2-i\right)}{1+i} \\ {z}_4=5+12i is \end{gathered}$$

$$\begin{gathered} z_1=(3-4i)(4+3i) \\ \Rightarrow \left|z_1\right|=\left|3-4i\right|\left|4+3i\right| \\ \Rightarrow \left|z_1\right|=\sqrt{9+16}\sqrt{16+9} \\ \Rightarrow \boxed{\left|z_1\right|=25} \end{gathered}$$

$z_2=\frac{3+4i}{1+i}\Rightarrow \left|z_2\right|=\frac{\left|3+4i\right|}{\left|1+i\right|}=\frac{\sqrt{9+16}}{\sqrt{1+1}}=\frac{5}{\sqrt{2}}$

$$\begin{gathered} z_3=\frac{\left(3+i\right)\left(2-i\right)}{1+i} \\ \left|z_3\right|=\frac{\left|\left(3+i\right)\right|\left|2-i\right|}{\left|1+i\right|}=\frac{\sqrt{9+1}\sqrt{4+1}}{\sqrt{1+1}}=5 \end{gathered}$$

$$\begin{gathered} z_4=5+12i \\ \Rightarrow \left|z_4\right|=\sqrt{25+144}=13 \end{gathered}$$