Answer
(a) T = 1.41 w
(b) $F_1 = w$
$F_2 = w$
Work Step by Step
The vertical component of the diagonal string is equal in magnitude to the weight $w$.
$T~sin(45^{\circ}) = w$
$T = \frac{w}{sin(45.0^{\circ})}$
$T = 1.41~w$
The tension in the diagonal string is 1.41 w.
(b) $F_2$ is equal in magnitude to the horizontal component of the tension in the string.
$F_2 = T~cos(45.0^{\circ})$
$F_2 = \frac{w}{sin(45.0^{\circ})}~cos(45.0^{\circ})$
$F_2 = w$
Similarly, $F_1$ is equal in magnitude to the horizontal component of the tension in the string. Therefore, $F_1$ is also equal to w.
