For the first part, note that any of the two equations use differentiation of $u$. If you add a constant to $u$ then when you differentiate, that constant vanishes, and therefore the equations still hold.
For the second part, note the following well known formulas from calculus; maybe they were presented in your pde course:
Integration by parts: $\int_\Omega \frac{\partial u}{\partial x_i}v d \Omega=\int_\Gamma uv\nu_i d \Gamma-\int_\Omega u \frac{\partial v}{\partial x_i} d \Omega,$ where $d\Omega$ is standard integration and $d \Gamma$ is surface integration. Summing this for $v=v_i$ each $i=1..n$ this leads to
$\int_\Omega \nabla u \cdot v d \Omega=\int_\Gamma (uv)\cdot \nu d \Gamma-\int_\Omega u ( \nabla \cdot v) d \Omega,$
and taking $v=\nabla v $ in the last formula you get Green's Formula. $ \int \nabla u \cdot \nabla v d\Omega =\int_\Gamma u \nabla v \cdot \mu d \Gamma-\int_\Omega u \Delta v d \Omega$
Taking $u\equiv 1$ in the second formula you get the divergence theorem which plainly states that $0 = \int_\Gamma v \cdot \nu d \Gamma-\int_\Omega \text{div v}d \Omega $
These formulas are basic, so you better keep them in mind. To solve point two, just note that putting $v=\nabla u$ in the last formula, or putting $u=1$ in Green's formula you get that
$0=\int_\Gamma \frac{\partial u}{\partial n}d \nu- \int_\Omega \Delta u d \Omega$
Now, since $\Delta u=f$ and $\frac{\partial u}{\partial n}=0$, replacing these two in the equality gets you to the result.
A nice video about the divergence theorem is presented here