Answer
(a) $ 4.90 \, \mathrm{m} $
(b) $ 60 \, \mathrm{N} $
Work Step by Step
(a) I choose to define positive torque as counterclockwise and select the hinge as the origin, thereby eliminating the unknown force it exerts on the pole (see diagram). Since the pole is in equilibrium, I know that the net torque is zero, and I can solve for $T_y$ and then the angle $\theta$:
$$ \Sigma \tau = T_yl - 0.5w_\mathrm{p}l - w_\mathrm{w}l = 0 \Rightarrow T_y = 0.5w_\mathrm{p} + w_\mathrm{w} \Rightarrow $$
$$ T\sin{\theta} = 0.5w_\mathrm{p} + w_\mathrm{w} \Rightarrow \theta = \arcsin \frac{0.5w_\mathrm{p} + w_\mathrm{w}}{T} = $$
$$ \arcsin \frac{0.5(200 \, \mathrm{N}) + 600 \, \mathrm{N}}{1000 \, \mathrm{N}} = 44.43^\circ $$
Now we can use this angle to find $h$:
$$ h = l \tan{\theta} = (5.00 \, \mathrm{m})\tan{44.43^\circ} = 4.90 \, \mathrm{m} $$
(b) Finding the new angle and then tension:
$$ \theta_2 = \arctan{\frac{4.90 \, \mathrm{m} - 0.50 \, \mathrm{m}}{5.00 \, \mathrm{m}}} = 41.35^\circ$$
$$ T_2 = \frac{0.5w_\mathrm{p} + w_\mathrm{w}}{\sin{\theta_2}} = \frac{0.5(200 \, \mathrm{N}) + 600 \, \mathrm{N}}{\sin{41.35^\circ}} = 1060 \, \mathrm{N} \Rightarrow$$
$$ \Delta T = 60 \, \mathrm{N} $$
