Answer
(a) $ F_1 = F_2 = 39.2 \, \mathrm{N} $, both directed up.
(b) $ F_1 = 60.0 \, \mathrm{N}, F_2 = 18.4 \, \mathrm{N} $, both directed up.
(c) $ F_1 = 164 \, \mathrm{N}$ up, $ F_2 = 85.9 \, \mathrm{N}$ down.
(d) $ \Omega = 0.094 \, \mathrm{rev/s}$
Work Step by Step
(a) $ F_1 = F_2 = mg/2 = (8.00 \, \mathrm{kg})(9.80 \, \mathrm{N/kg})/2 = 39.2 \, \mathrm{N}$. Both forces are directed upward.
(b) With positive $y$-direction up and postive torque clockwise (as illustrated in picture), we can determine the net torque about the first hand (left in picture) which is $ \Sigma \tau = mgR - 2F_2R = R(mg - 2F_2)$. Inserting this into the precession speed equation and solving for $F_2$:
$$ \Omega = \frac{\Sigma \tau}{I\omega} = \frac{R(mg - 2F_2)}{mr^2\omega} \Rightarrow F_2 = \frac{m}{2} \bigg( g - \frac{\Omega \omega r^2}{R} \bigg) = $$
$$ \frac{8.00 \, \mathrm{kg}}{2}\bigg( 9.80 \, \mathrm{N/kg} - \frac{(0.050 \times 2\pi \, \mathrm{rad/s})(5.00 \times 2\pi \, \mathrm{rad/s}) (0.325 \, \mathrm{m})^2}{0.200 \, \mathrm{m}} \bigg) = 18.4 \, \mathrm{N} $$
Since $F_1 + F_2 = mg$ we can easily find $F_1$ that way:
$$ F_1 = (8.00 \, \mathrm{kg})(9.80 \, \mathrm{N/kg}) - 18.4 \, \mathrm{N} = 60.0 \, \mathrm{N} $$
Both forces are in the positive $y$-direction (upwards).
(c) We'll find the forces in the same way we did in (b):
$$ F_2 = \frac{8.00 \, \mathrm{kg}}{2}\bigg( 9.80 \, \mathrm{N/kg} - \frac{(0.300 \times 2\pi \, \mathrm{rad/s})(5.00 \times 2\pi \, \mathrm{rad/s}) (0.325 \, \mathrm{m})^2}{0.200 \, \mathrm{m}} \bigg) = -85.9 \, \mathrm{N} $$
$$ F_1 = 164 \, \mathrm{N} $$
$F_1$ is directed upward, $F_2$ downward.
(Note that the book has the answer to part (c) wrong, it claims that the forces $F_2$ and $F_1$ are $-86.2 \, \mathrm{N}$ and $165 \, \mathrm{N}$, respectively.)
(d) Now we simply solve for $\Omega$ when $ \Sigma \tau = mgR$:
$$\Omega = \frac{mgR}{mr^2\omega} = \frac{gR}{r^2\omega} = \frac{(9.80 \, \mathrm{m/s^2})(0.200 \, \mathrm{m})}{(0.325 \, \mathrm{m})^2(5.00 \times 2 \pi \, \mathrm{rad/s})} = 0.591 \, \mathrm{rad/s} = 0.094 \, \mathrm{rev/s}$$
