Answer
(a) $ 3.41 \, \mathrm{kN} $
(b) $ 3.41 \, \mathrm{kN} $ and $ 7.60 \, \mathrm{kN} $
Work Step by Step
(a) I'll call the angle $\phi$, the tension in the guy wire $T$, the force exerted on the lower boom $N$, the length of the boom $l$ and the weights $w_\mathrm{box}$ and $w_\mathrm{boom}$. With positive torque clockwise and origin at the base of the boom, we have the following relationship for rotational equilibrium:
$$ \Sigma \tau = w_\mathrm{box}l\cos{\phi} + 0.35w_\mathrm{boom}l\cos{\phi} - Tl\sin{\phi} = 0$$
Solving for $T$ yields:
$$ T = \frac{w_\mathrm{box}\cos{\phi} + 0.35w_\mathrm{boom}\cos{\phi}}{\sin{\phi}} = \frac{w_\mathrm{box} + 0.35w_\mathrm{boom}}{\tan{\phi}} = $$
$$ \frac{(5000 \, \mathrm{N}) + 0.35(2600 \, \mathrm{N})}{\tan{60.0^\circ}} = 3410 \, \mathrm{N} = 3.41 \, \mathrm{kN} $$
(b) The boom is in translational equilibrium as well. Thus, with positive $y$-direction up and postive $x$-direction to the right:
$$ \Sigma F_y = N_y - w_\mathrm{box} - w_\mathrm{boom} = 0 $$
$$ \Sigma F_x = N_x - T = 0 $$
Solving for our targets $N_y$ and $N_x$:
$$ N_y = 5000 \, \mathrm{N} + 2600 \, \mathrm{N} = 7600 \, \mathrm{N} = 7.60 \, \mathrm{kN} $$
$$ N_x = 3410 \, \mathrm{N} = 3.41 \, \mathrm{kN} $$
