I would expect many of your graphs to not have distinct eigenvalues.
Here's one reason: They have a very large number of degree $1$ vertices. Enough, in fact, that there will probably be a decent number of degree $1$ vertices which share a common neighbor with another degree $1$ vertex. These will give rise to many independent vectors in the nullspace of your matrix, so $0$ will be a multiple eigenvalue.
More generally, you can imagine the following situation: We say a subgraph $G'$ "dangles" from a vertex $v$ if the only edge connecting $G'$ to the remainder of the graph goes from some vertex $w \in G'$ to $v$. If you have two copies of $G'$ dangling from the same $v$ in the same way ($w$ occupies the same position in both graphs), then all the eigenvalues of $G'$ appear in the spectrum of $G$: The eigenvector is equal to $x$ on one copy of $G'$, $-x$ on the other copy, and $0$ elsewhere in the graph, where $x$ is an eigenvector of $G'$.
You can check, for example, that if instead of a power law graph you had the Erdős-Rényi graph $G(n,c/n)$ for fixed $c$, then every tree $T$ of fixed size appears as a dangling subgraph $c'n$ times (where $c'$ is positive and depends on the tree and $c$). This in turn implies there will be on average $c''n$ vertices having two copies of $T$ dangling from them, and for every such vertex you'll get a copy of all the eigenvalues of $T$. You might expect a similar phenomenon to happen with power law graphs, though I haven't checked this. (I did check a couple of graphs from the Stanford collection to see if they had $0$ as a multiple eigenvalue, and the ones I checked had it many times over).
This is all dependent on there being a sizable number of degree one vertices though. If we move to, say, random Erdős-Rényi graphs with edge probability $1/2$, I would expect the eigenvalues to be distinct with high probability, though as far as I know proving this is still an open problem (EDIT: This was recently proved by Tao and Vu -- their preprint is here and is described in this blog post of Tao's).
For more on this issue, albeit for random preferential attachment trees instead of graphs, check out the recent paper Spectra of Large Random Trees by Bhamidi, Evans, and Sen.
The probability of picking a certain node is the probability of picking one of its adjacent edges times the probability of choosing its end:
$$\frac{1}{2}\frac{d}{e},$$
where $d$ is its degree of the node and $e$ the number of links.
The probability of choosing any node of degree $d$ is the sum of these probabilities:
$$n(d) \frac{d}{2e},$$
where $n(d)$ is the number of nodes of degree $d$.
Now observe that $P(d) = n(d)/N$ and that $E(d) = 2e/N$, where $N$ is the number of nodes.
With regards to Q2, consider as a counter example a network with three nodes and two links. The probability of having degree 2 is 1/3, but the probability of choosing the node with degree 2 by this process is 1/2.
Best Answer
@bof hint pointed in the right direction. If both maximum degree and diameter are bounded you can find an upper bound for the number of nodes in any network of the sequence by doing a BFS from any node. With maximum degree $d$ and maximum diameter $k$ the total number of nodes can be at most the Moore bound: $$ n \leq 1 + d \sum_{i=0}^{k-1}(d-1)^{i}$$