Hint 3

The station has an axis of symmetry and so $I_1=I_2$. For a constant torque about the (body-fixed) $x_2$-axis, Euler's equations are

$$I_1\dot\omega_1+(I_3-I_1)\omega_2\omega_3 = 0$$

$$I_2\dot\omega_2+(I_1-I_3)\omega_3\omega_1 = \tau$$

$$I_3\dot\omega_3 = 0$$

The third equation implies that

$$\omega_3=\hbox{\rm constant}$$

The first two equations can be written

$$\dot\omega_1+\Omega\omega_2 = 0$$

$$\dot\omega_2-\Omega\omega_1 = \tau/I_1$$

where

$$\Omega={I_3-I_1\over I_1}\omega_3$$

These can be solved either by differentiating the second with respect to $t$and using the first to eliminate $\dot\omega_1$ or by introducing the complex variable $q=\omega_1+{\rm i}\omega_2$. The general solution is

$$q(t)=-{\tau\over I_1\Omega}+Ae^{{\rm i}\Omega t}$$

Imposing the initial condition that $\omega_1(0)=\omega_2(0)=0$ fixes the constant

$$A={\tau\over I_1\Omega}$$

Taking real and imaginary parts of $q$ gives the forms for $\omega_{1,2}(t)$in the question.

At $t=\pi/\Omega$ the angular velocity is

$$\hbox{\boldmath $\omega$}=\left(-{2\tau\over I_1\Omega},0,\omega_3\right)$$

Since our axes lie along the principal axes, the angular momentum is just

$${\bf L}=\left(-{2\tau\over \Omega},0,I_3\omega_3\right)$$