Let's break down the problem step by step. The vector field given is:

$\mathbf{F} = \langle 12x^3 - 36xy^2, 12y^3 - 36x^2y \rangle$

We are asked to:

a. Verify that the curl and divergence of the given field are zero.

Curl in two dimensions (2D): The curl of a vector field $\mathbf{F}(x, y) = \langle P(x, y), Q(x, y) \rangle$ in 2D is given by:

$\text{Curl}(\mathbf{F}) = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$

Divergence: The divergence of a vector field $\mathbf{F}(x, y) = \langle P(x, y), Q(x, y) \rangle$ is given by:

$\text{Div}(\mathbf{F}) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y}$

Let's compute these derivatives.

b. Find a potential function $\phi$ and a stream function $\psi$ for the field.

We need to find two functions:

1. $\phi$ such that $\mathbf{F} = \nabla \phi$, meaning $\frac{\partial \phi}{\partial x} = P$ and $\frac{\partial \phi}{\partial y} = Q$.
2. $\psi$ such that the field $\mathbf{F}$ is perpendicular to the gradient of $\psi$, i.e., $\frac{\partial \psi}{\partial y} = P$ and $-\frac{\partial \psi}{\partial x} = Q$.

c. Verify that $\phi$ and $\psi$ satisfy Laplace's equation.

Laplace's equation is given by:

$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = 0 \quad \text{and} \quad \frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2} = 0$

Now, let's proceed by calculating these steps. I'll begin by computing the curl and divergence of the vector field.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can guide you through the process of calculating the curl and divergence manually.

If you'd like, I can walk you through the rest of the steps by hand. Would you like that?