#### Answer

(a) $d_0 = 8.5~ly$
$t_0 = 17~years$
(b) $d = 7.36~ly$
$t = 14.72~years$
(c) Both sets of answers are correct according to each frame of reference.

#### Work Step by Step

(a) We can find the distance according to earthlings:
$d_0 = (2)(4.25~ly) = 8.5~ly$
We can find the time of the journey according to earthlings:
$t_0 = \frac{d}{v} = \frac{8.5~ly}{0.500~c} = 17~years$
(b) We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{(0.500~c)^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{0.75}}$
$\gamma = 1.155$
We can find the distance according to the crew:
$d = \frac{d_0}{\gamma}$
$d = \frac{8.5~ly}{1.155}$
$d = 7.36~ly$
We can find the time of the journey according to the crew:
$t = \frac{t_0}{\gamma}$
$t = \frac{17~years}{1.155}$
$t = 14.72~years$
(c) Both sets of answers are correct according to each frame of reference.