Answer
(a) $11m$
(b) Less than
(c) $6.1m$
Work Step by Step
(a) We know that
$v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
We plug in the known values to obtain:
$v=\frac{0.75c+(-0.40c)}{1+\frac{(0.75c)(-0.40c)}{c^2}}$
$\implies v_c=0.5c$
Now $L=1.3m\sqrt{1-\frac{(0.5c)^2}{c^2}}$
$\implies L=11m$
(b) We know that if $v$ is greater then $\sqrt{1-\frac{v^2}{c^2}}$ should be less, which means that the value of $L_{\circ}\sqrt{1-\frac{v^2}{c^2}}$ should be less . Thus, we conclude that the length measured by an observer on Earth would be less than the value which is found in part(a).
(c) We know that
$v=\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}$
We plug in the known values to obtain:
$v=\frac{0.40c+0.75c}{1+\frac{(0.40c)(0.75c)}{c^2}}$
$\implies v=0.8846c$
and $L=(13m)\sqrt{1-\frac{(0.8846c)^2}{c^2}}$
$\implies L=6.1m$
