Bellman-Ford Termination when there is no change on vertex weights?
Part (1) is indeed true. The idea is that in $k$ steps the algorithm is only considering paths of length $k$. Indeed, you can prove by induction on $k$ that the path realizing the value of $d(v)$ after $k$ steps is at most $k$ edges long. 

If the algorithm terminates after $k$ steps, then the results of the first $k-1$ steps are already optimal, hence (1).

Part (2) is indeed false, and I'm sure you can give a counterexample on your own.



The simultaneous version of the Bellman&ndash;Ford algorithm, referred to in part (1), has the following pseudo-code (where $d(x,y)$ is the weight of the edge connecting $x$ and $y$):

A[0,x,y] = ∞ for all x,y
t ← 0
do
  for all x,y:
    A[t+1,x,y] = min(A[t,x,y], { A[t,x,z] + d(z,y) : all z ≠ x,y })
  t ← t+1
while A[t,x,y] ≠ A[t-1,x,y] for some x,y
return A[t,x,y]


Take as an example the (undirected) graph $x - y - z - w$ (with unit weights). Here is a trace of the algorithm:

$$
\begin{array}{c|cccccc}
t &amp; A[t,x,y] &amp; A[t,x,z] &amp; A[t,x,w] &amp; A[t,y,z] &amp; A[t,y,w] &amp; A[t,z,w] \\\hline
0 &amp; \infty &amp; \infty &amp; \infty &amp; \infty &amp; \infty &amp; \infty \\
1 &amp; 1 &amp; \infty &amp; \infty &amp; 1 &amp; \infty &amp; 1 \\
2 &amp; 1 &amp; 2 &amp; \infty &amp; 1 &amp; 2 &amp; 1 \\
3 &amp; 1 &amp; 2 &amp; 3 &amp; 1 &amp; 2 &amp; 1 \\
4 &amp; 1 &amp; 2 &amp; 3 &amp; 1 &amp; 2 &amp; 1
\end{array}
$$