See atach file
Question 2 was asked by another user and was answered here
personally, i would have used Soroban's method to do it...but CaptainBlack's method was so cool! and it seems to have much wider applications. I don't know what Wallis' formula is though, so you should be able to make a better choice than i can
Here's question 1
Recall that for a vector field, $\displaystyle \bold{F} = P \bold {i} + Q \bold {j} + R \bold {k}$, $\displaystyle \mbox{curl} \bold{F}$ is given by:
$\displaystyle \mbox{curl} \bold{F} = \nabla \times \bold{F}$
where $\displaystyle \times$ means the cross-product, and $\displaystyle \nabla = \frac {\partial}{\partial x} + \frac {\partial}{\partial y} + \frac {\partial}{\partial z}$
That is, $\displaystyle \mbox {curl} \bold{F} = \left| \begin{array}{ccc} \bold{i}&\bold{j}&\bold{k} \\ \partial / \partial x & \partial / \partial y & \partial / \partial z \\ P & Q & R \end{array} \right|$ $\displaystyle = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\bold{i}+ \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) \bold{j} + \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \bold{k} $
Here we have $\displaystyle \bold {F} = z e^{xy} \bold{i} + 2x^2 z \sin y \bold{j} + \left( 2x^2 + 3y^2 \right) \bold{k}$
and we want the curl at $\displaystyle (2,3,2)$
So, $\displaystyle \mbox{curl} \bold {F} = \left( \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial y} - \frac{\partial \left( 2x^2 z \sin y \right)}{\partial z}\right)\bold{i}+ $ $\displaystyle \left( \frac{\partial \left(z e^{xy} \right)}{\partial z} - \frac{\partial \left( 2x^2 + 3y^2 \right)}{\partial x} \right) \bold{j} + \left( \frac{\partial \left( 2x^2 z \sin y \right)}{\partial x} - \frac{\partial \left( z e^{xy}\right)}{\partial y} \right) \bold{k}$
..............$\displaystyle = \left( 6y - 2x^2 \sin y \right) \bold{i} + \left( e^{xy} - 4x \right) \bold{j} + \left( 4x z \sin y - xz e^{xy} \right) \bold {k}$
Since we want the curl at $\displaystyle (2,3,2)$, we simply plug in $\displaystyle x = 2 \mbox { , } y = 3 \mbox { , and } z = 2 $. I leave that to you