I will answer your question 1 as I don't think I have enough information to answer the other ones.
Short answer:
If $\nabla\cdot\boldsymbol u=0$, then $\nabla\cdot\big(\nabla\boldsymbol u+(\nabla\boldsymbol u)^\intercal\big)=\nabla^2 \boldsymbol u$.
LONG ANSWER:
(This is partially for my own reference, as well as others hopefully.)
The complete, general Navier-Stokes equations are as follows:
$$\rho\frac{\mathrm D\boldsymbol u}{\mathrm Dt}=\nabla\cdot\boldsymbol \sigma +\boldsymbol F \\ \frac{\mathrm D\rho}{\mathrm Dt}=-\rho\nabla\cdot\boldsymbol u$$
The first equation is the "momentum equation" and the second is the "continuity equation".
The symbols mean:
$\frac{\mathrm D}{\mathrm Dt}=\frac{\partial}{\partial t}+\boldsymbol u\cdot \nabla$ : The "material derivative"
$\boldsymbol u$ : The velocity of the fluid
$P$ : the pressure
$\rho$ : the density
$\boldsymbol F$ : An external force (e.g gravity)
$\boldsymbol \sigma$ : The Cauchy stress tensor.
The first equation describes conservation of momentum and the second describes conservation of mass.
The first assumption is that the Cauchy stress tensor takes the form
$$\boldsymbol \sigma=-P\mathbf g+\boldsymbol \tau$$
Where $\mathbf g$ is the metric tensor and $\boldsymbol \tau$ is the viscous stress tensor. The motivation for this is that when there is no viscous stress, the total stress in the fluid should be equal to the hydrostatic stress $-P\mathbf g$.
The second assumption is that we are working at a sufficiently large scale, so that the Knudsen number is quite close to zero. This in turn implies that the infinitesimal torque around any individual point is zero, which in turn implies that $\boldsymbol \sigma$ is a symmetric tensor. Since $\mathbf g$ is symmetric this implies $\boldsymbol \tau$ is symmetric as well.
The third assumption is the Newtonian hypothesis, which is that the viscosity depends linearly on the strain rate $S_{ij}=(\nabla u)_{(ij)}=\frac{1}{2}(\nabla u)_{ij}+\frac{1}{2}(\nabla u)_{ji}$ :
$$\tau^{ij}=C^{ijkl}S_{kl}$$
Due to the symmetries in the various tensors involved and under the assumption that the fluid is isotropic, i.e looks the same in all directions, we can reduce these $81$ possible coefficients to only two, the first coefficient of viscosity or the "dynamic viscosity" $\mu$ and the second viscosity $\lambda$. (See here for some references on the topic.) In full, we write
$$\boldsymbol \tau = \lambda S^k{}_k \mathbf g+2\mu\mathbf S$$
Where $S^k{}_k=\nabla\cdot \boldsymbol u$ is the trace of the strain rate tensor, also known as the dilation rate.
Finally, the Stokes hypothesis is that $\operatorname{tr}\boldsymbol\tau=0$. Because the trace of a matrix is the sum of its eigenvalues, if the trace were nonzero it would mean that the viscosity of the fluid is pushing the fluid in a certain direction. But, we understand viscosity to simply be a realization of the internal friction of the fluid, and should not "push" the fluid in any particular direction. We find that
$$\operatorname{tr}\boldsymbol \tau=\tau^i{}_i=\lambda S^k{}_k \underbrace{g^i{}_i}_{=3}+2\mu S^k{}_k \\ =S^k{}_k(3\lambda +2\mu)$$
So the Stokes hypothesis is that the $3\lambda+2\mu=0$, or $\lambda=-2\mu/3$. Note that this assumption is not always valid, but it is a good enough approximation in most cases.
So, the momentum equation is then
$$\rho \frac{\mathrm D\boldsymbol u}{\mathrm Dt}=\nabla\cdot \boldsymbol \sigma+\boldsymbol F$$
Let's assume that there is no external force, and use the form of the viscous stress tensor found earlier. We get
$$\rho\frac{\mathrm D\boldsymbol u}{\mathrm Dt}=\nabla\cdot(-P\mathbf g+\boldsymbol \tau) \\ =-\nabla P+\nabla\cdot \left(\frac{-2}{3}\mu S^k{}_k\mathbf g+2\mu\mathbf S\right) \\ =-\nabla P -\frac{2}{3}\mu\nabla(\nabla\cdot\boldsymbol \mu)+\mu \nabla\cdot\big(\nabla\boldsymbol u+(\nabla\boldsymbol u)^\intercal\big)$$
We also rewrite the continuity equation in the form $\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol u)=0$ (exercise!)
Finally, we reach the most common form of the Navier-Stokes equations you will see
$$\boxed{\rho\frac{\mathrm D\boldsymbol u}{\mathrm Dt}=-\nabla P+\mu\nabla^2\boldsymbol u+\frac{1}{3}\mu\nabla(\nabla\cdot \boldsymbol u)\\ \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\boldsymbol u)=0}$$
But we can go further. If we now assume a incompressible, homogeneous fluid ($\rho=$constant) we can introduce the "kinematic pressure" $p=P/\rho$ and the "kinematic viscosity" and write the \textit{incompressible Navier-Stokes equations}:
$$\boxed{\frac{\mathrm D\boldsymbol u}{\mathrm Dt}=-\nabla p+\nu\nabla^2\boldsymbol u \\ \nabla\cdot\boldsymbol u=0}$$
Finally, we aim to obtain the equation for Stokes flow. We take the above incompressible Navier-Stokes equations and introduce some reference values for the velocity and distance called $U,L$ respectively. If we have chosen these references values correctly in our application, the dimensionless parameter $\operatorname{Re}=\frac{UL}{\nu}$ called the Reynolds number will roughly describe the ratio of the inertial ($\rho \boldsymbol u\cdot \nabla\boldsymbol u $) to the viscous ($\mu\nabla^2\boldsymbol u$) forces. Next we introduce the non dimensional variables
$$\bar{t}=\frac{tU}{L} \implies \frac{\partial}{\partial t}=\frac{U}{L}\frac{\partial}{\partial \bar{t}}\\ \bar{\boldsymbol x}=\frac{\boldsymbol x}{L}\implies \nabla=\frac{1}{L}\bar{\nabla} \\ \bar{\boldsymbol u}=\frac{\boldsymbol u}{U} \\ \bar{p}=\frac{pL}{\nu U}$$
You will find (exercise!) that our above incompressible momentum equation can be written in terms of these non dimensional variables as
$$\operatorname{Re}\cdot\left(\frac{\partial \bar{\boldsymbol u}}{\partial\bar t}+\bar{\boldsymbol u}\cdot \bar{\nabla}\bar{\boldsymbol u}\right)=-\bar{\nabla}\bar{p}+\bar{\nabla}^2\bar{\boldsymbol u}$$
In the viscous limit $\operatorname{Re}\to 0$ we recover the Stokes equation $\nabla^2\boldsymbol u=\nabla p$.
