Using Laplace in this circuit, how to find part C(Vx)?
You used the correct Laplace equivalences for the elements, but the transfer function seems wrong in the numerator. First, the equivalent impedance seen from the left to right, is the resistor parallel with the inductance, \$Z_{RL}\$, and the impedance seen from the right to left is the resistor parallel with the capacitor, \$Z_{RC}\$ (see the picture below). Now the transfer function becomes:
$$\begin{align}
Z_{RL}&amp;=\dfrac{4s}{4s+1} \\
Z_{RC}&amp;=\dfrac{2}{s+2} \\
V_x&amp;=\dfrac{5}{s}\left(\dfrac{Z_{RL}}{\dfrac{2}{s}+Z_{RL}}+\dfrac{Z_{RC}}{4s+Z_{RC}}\right)&amp;=\dfrac{1}{s}\cdot\dfrac{10s^2+5}{2s^2+4s+1} \tag{1}
\end{align}$$
If \$V_x\$ is related to a unity input, \$V_x=H(s)\$. To verify this:

The sources are I1 and C1, which form an integrator, and together with the value ac 5 form the Laplace equivalent of \$\dfrac{5}{s}\$.
For the c) part you have two ways you can solve this: inverse Laplace, or differential equations. Since you already have the transfer function, it might be easier to apply the inverse Laplace. Might not. Here's one way to do it. First, expand into partial fractions:
$$\begin{align}
(1)&amp;\rightarrow\dfrac{5}{s}-\dfrac{20}{2s^2+4s+1}&amp;=U(s)-A(s) \tag{2} \\
\mathscr{L}^{-1}\{U(s)\}&amp;=5 \tag{3}
\end{align}$$
The roots for the denominator of the second term, \$A(s)\$, are real roots, \$\{-0.293, -1.707\}\$, therefore it's an overdamped case, with the inverse Laplace:
$$\mathscr{L}^{-1}\{A(s)\}=10\sqrt{2}\sinh\left(\dfrac{t}{\sqrt{2}}\right)\mathrm{e}^{-t} \tag{4}$$
Combine (3) and (4) to give the resulting response:
$$h(t)=5-10\sqrt{2}\sinh\left(\dfrac{t}{\sqrt{2}}\right)\mathrm{e}^{-t} \tag{6}$$
However, the verification seems a bit awkward. Building the time domain circuit shows slightly different results:

OTOH, using the IFFT on the frequency response from the 1st picture shows a match, though LTspice considers the electrical, rather than the mathematical equivalent (hence the time correction):