A spool of mass 1.2 kg rests on a rough horizontal floor on its two outer rims, of radius R = 0.15…

Step 1. String from the bottom of the axle. Its line is $R - r = 0.10\ \text{m}$ above the floor. $\tau_P = 3.0\ \text{N} \times 0.10\ \text{m} = 0.30\ \text{N m}$, so $\alpha = \dfrac{0.30}{0.036} = 8.33\ \text{rad/s}^2$ and $a = \alpha R = 1.25\ \text{m/s}^2$, in the direction of the pull.

Friction: $T + f = ma$, so $f = 1.5\ \text{N} - 3.0\ \text{N} = -1.5\ \text{N}$. That is $1.5\ \text{N}$ against the pull.

Step 2. String from the top of the axle. Its line is $R + r = 0.20\ \text{m}$ up. $\tau_P = 0.60\ \text{N m}$, $\alpha = 16.67\ \text{rad/s}^2$, $a = 2.5\ \text{m/s}^2$. Friction: $f = ma - T = 3.0\ \text{N} - 3.0\ \text{N} = 0$. No friction is needed at all: about the centre, the string's torque $Tr = 0.15\ \text{N m}$ is exactly $I\alpha$ for this rolling.

Step 3. Pulled at an angle. The string leaves the axle along a tangent, so its line is always a distance $r$ from the centre. Its torque about P is $T(R\cos\theta - r)$. It is zero when the line passes through P: $\cos\theta = \dfrac{r}{R} = 0.333$, $\theta = 70.5^\circ$. For smaller angles the spool rolls towards the pull; for larger ones it rolls away from the pull.