To find the torque in shaft segment $\left(1\right)$, draw a free-body diagram $\left(\mathrm{FBD}\right)$ that cuts through segment $\left(1\right)$ and includes the free end of the shaft.

Sum the torques acting on the $\mathrm{FBD}$:

$$\begin{gathered} \sum _{}\mathrm{M}=-{\mathrm{T}}_1+600 \mathrm{N}-\mathrm{m} -200 \mathrm{N}-\mathrm{m}=0 \\ \therefore {\mathrm{T}}_1=400 \mathrm{N}-\mathrm{m} \end{gathered}$$

Plot this torque on the torque diagram.