For first $5\, sec$ motion ${s_5} = 10\;metre$

$s = ut + \frac{1}{2}a{t^2} \Rightarrow 10 = 5u + \frac{1}{2}a{(5)^2}$

$2u + 5a = 4$ …(i)

For first $8 \,sec$ of motion ${s_8} = 20\;metre$

$20 = 8u + \frac{1}{2}a{(8)^2} \Rightarrow 2u + 8a = 5$ …(ii)

By solving $u = \frac{7}{6}m/s\;{\rm{and }}a = \frac{1}{3}m/{s^2}$

Now distance travelled by particle in Total $10$ sec.

${s_{10}} = u \times 10 + \frac{1}{2}a{(10)^2}$

By substituting the value of $u$ and $a$ we will get ${s_{10}} = 28.3\;m$

so the distance in last $2\;\sec = {s_{10}} - {s_8}$

$ = 28.3 - 20 = 8.3\,m$