From a free-body diagram that cuts through shaft $\left(2\right)$ and includes gear $\mathrm{D}$, the following equilibrium equation shows ${\mathrm{T}}_2$ to have a positive internal torque:

$\sum _{}{\mathrm{M}}_{{\mathrm{x}}^{\text{'}}}={\mathrm{T}}_{\mathrm{D}}-{\mathrm{T}}_2=0 \therefore {\mathrm{T}}_2={\mathrm{T}}_{\mathrm{D}}=+131.303 \mathrm{lb}-\mathrm{ft}$

The presence of gears $\mathrm{B}$ and $\mathrm{C}$ will cause shaft $\left(1\right)$ to rotate opposite to the direction of shaft $\left(2\right)$. The sign of the torque in shaft $\left(2\right)$ is positive. Therefore, the sign of ${\mathrm{T}}_1$ must be negative:

${\mathrm{T}}_1=-43.768 \mathrm{lb}-\mathrm{ft}$