Once $k>2$ the answer to #1 depends on the graph;
even in the limiting determinstic case of $p=1$, and taking $n_0=1$,
the expected time till full zombification can range from about $C \log n$
(for an expander graph) to at least $n/k$ (when $k$ is even, the node set is
${\bf Z}/n{\bf Z}$, and each node is joined to its $k$ nearest neighbors).

As to #2: for large $t$, the probability that there is still an
uninfected node $-$ which is essentially the difference between $n$
and the expected value of $z(t)$ $-$ decays exponentially with $t$.
The base of the exponent is $(1-p)^{b_{\min}}$, where $b_{\min}$
is the minimum number $b(S)$ of edges connecting $S$ and its complement,
and $S$ ranges over all nonempty sets of nodes that are initially uninfected
and might become the set of yet-uninfected nodes at some point.
(This assumes of course that $n_0<n$: if $n_0=n$ then $z(t)=n$
for all $n$.) Usually $b=k$, realized by singleton sets;
but $b$ can be smaller if the graph has a bottleneck,
e.g. for this cubic graph

_{(source: harvard.edu)}

the base is $1-p$ as long as the zombies were initially on
just one side of the graph.

To obtain the $(1-p)^{b_{\min}}$ formula,
regard the outbreak as a dynamical system on the
subsets $S$ of the set $S_0$ of initially uninfected nodes.
At each $t$ the system's state is the set of nodes
not yet infected by time $t$.
The probability that $S$ goes to itself is $(1-p)^{b(S)}$,
and otherwise $S$ goes to a proper subset. Thus the
dynamical system is triangular (with respect to the
partial order of set inclusion), with eigenvalues equal to
the diagonal entries $(1-p)^{b(S)}$.
The dominant eigenvalue of $1$ occurs only for $S=\emptyset$
because the graph is connected. The next eigenvalue is
$(1-p)^{b_{\min}}$, and occurs for all nonempty $S$ minimizing $b(S)$;
these are usually but not always the singletons in $S_0$.
(Sometimes not every $S \subset S_0$ can be reached from $S_0$, but clearly
at one singleton is reachable unless $n_0=n$.)