Converting the circuit into it's Laplace transform

Here, $V_b=V_2$

& $I_b=\dfrac{V_2}{\dfrac{2}s}=\dfrac{sV_2}2$

$V_a=2sI_b+V_2=2s\left(\dfrac{sV_2}{2}\right)+V_2=(s^2+1)V_2$

$I_1=\dfrac{V_a}{\dfrac{1}s}+I_b=sV_a+I_b=\left[s(s^2+1)V_2\right]+\dfrac{sV_2}2=\left[s(s^2+1)+\dfrac{s}2\right]V_2=\left[s^3+s+\dfrac{s}2\right]V_2=\left[s^3+\dfrac{3s}2\right]V_2$

$V_1=sI_1+V_a$

$V_1=s\left[s^3+\dfrac{3s}2\right]V_2+(s^2+1)V_2=\left(s\left[s^3+\dfrac{3s}2\right]+(s^2+1)\right)V_2$

Now, $\dfrac{V_1}{I_1}=\dfrac{\left(s\left[s^3+\dfrac{3s}2\right]+(s^2+1)\right)}{\left[s^3+\dfrac{3s}2\right]}$

$\therefore \underline{\underline{\dfrac{V_1}{I_1}=\dfrac{2s^4+5s^2+2}{2s^3+3s}}}$

$\dfrac{V_2}{I_1}=\dfrac{1}{s^3+\dfrac{3s}2}$

$\therefore \underline{\underline{\dfrac{V_2}{I_1}=\dfrac{2}{2s^3+3s}}}$