The period of a particle in SHM is 8 sec. At t = 0, it is at the mean position. The ratio  of the distances travelled by it in the first one second and in the 2^nd one second is

At t = 0, let particle move in +x direction. So, equation of SHM is :  $x=A\sin \omega t$
Now,  $\omega =\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}$  So,  $x=A\sin \left(\frac{\pi t}{4}\right)$
$v=\frac{A\pi }{4}\cos \left(\frac{\pi t}{4}\right)$ 
Distance travelled in t seconds,  $d=\underset{0}{\overset{t}{\int }}\left|v\right|dt$
$When\left[0<\left(\frac{\pi t}{4}\right)<\frac{\pi }{2}\right]or\left[0<t<2\sec \right]$       $v=\frac{A\pi }{4}\cos \left(0<\theta <\frac{\pi }{2}\right)\Rightarrow v=+ve$            
So, velocity is always positive over the time duration 2 seconds studied.
Distance travelled in first 1 second = 
$d_1=\underset{0}{\overset{1}{\int }}\frac{A\pi }{4}\cos \left(\frac{\pi t}{4}\right)dt={\left.\frac{A\pi }{4}\frac{4}{\pi }\sin \left(\frac{\pi t}{4}\right)\right|}_0^1=A\left(\frac{1}{\sqrt{2}}\right)$ 
Distance travelled in second 1 second = 
$d_2=\underset{1}{\overset{2}{\int }}\frac{A\pi }{4}\cos \left(\frac{\pi t}{4}\right)dt={\left.\frac{A\pi }{4}\frac{4}{\pi }\sin \left(\frac{\pi t}{4}\right)\right|}_1^2=A\left(1-\frac{1}{\sqrt{2}}\right)$

$\frac{d_1}{d_2}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$