I'll consider the "king" case, without the ability to "pass"; the other cases can be resolved similarly. Let $S_{\text{king}}(n,p)$ be the number of $n$-digit numbers with the king ending at $p \in \{0,1,\ldots,9\}$.
We have:
Boundary conditions: $S_{\text{king}}(1,p)=1$ for all $p \in \{0,1,\ldots,9\}$. (I'll assume we can start wherever we want.)
Recurrence relations:
$S_{\text{king}}(n,1)=\sum_{p \in \{2,4,5\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,2)=\sum_{p \in \{1,3,4,5,6\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,3)=\sum_{p \in \{2,5,6\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,4)=\sum_{p \in \{1,2,5,7,8\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,5)=\sum_{p \in \{1,2,3,4,6,7,8,9\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,6)=\sum_{p \in \{2,3,5,8,9\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,7)=\sum_{p \in \{0,4,5,8\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,8)=\sum_{p \in \{0,4,5,6,7,9\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,9)=\sum_{p \in \{0,5,6,8\}} S_{\text{king}}(n,p)$,
$S_{\text{king}}(n,0)=\sum_{p \in \{7,8,9\}} S_{\text{king}}(n,p)$.
This gives:
$$\scriptsize \begin{array}{r|rrrrrrrrrr|r}
& p=1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0 & \text{sum} \\
\hline
n=1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 10 \\
2 & 3 & 5 & 3 & 5 & 8 & 5 & 4 & 6 & 4 & 3 & 46 \\
3 & 18 & 24 & 18 & 26 & 35 & 26 & 22 & 29 & 22 & 14 & 234 \\
4 & 85 & 123 & 85 & 128 & 185 & 128 & 104 & 145 & 104 & 73 & 1160 \\
5 & 436 & 611 & 436 & 642 & 902 & 642 & 531 & 722 & 531 & 353 & 5806 \\
6 & 2155 & 3058 & 2155 & 3202 & 4551 & 3202 & 2619 & 3601 & 2619 & 1784 & 28946 \\
7 & 10811 & 15265 & 10811 & 15984 & 22611 & 15984 & 13138 & 17977 & 13138 & 8839 & 144558 \\
8 & 53860 & 76201 & 53860 & 79802 & 113108 & 79802 & 65411 & 89694 & 65411 & 44253 & 721402 \\
9 & 269111 & 380432 & 269111 & 398274 & 564041 & 398274 & 326857 & 447787 & 326857 & 220516 & 3601260 \\
10 & 1342747 & 1898811 & 1342747 & 1988228 & 2816703 & 1988228 & 1630618 & 2234819 & 1630618 & 1101501 & 17975020 \\
\end{array}$$
So there are $17975020$ phone numbers with $10$-digits that can be dialled by a king.
(There are some "sanity checks" we can apply to the above numbers. E.g. $S(n,1)=S(n,3)$ by symmetry.)
The above numbers were computed in GAP using the following code:
M:=[[1,1,1,1,1,1,1,1,1,1]]; # index 10 represents 0
for n in [2..10] do
M[n]:=[];
M[n][1]:=Sum([2,4,5],p->M[n-1][p]);
M[n][2]:=Sum([1,3,4,5,6],p->M[n-1][p]);
M[n][3]:=Sum([2,5,6],p->M[n-1][p]);
M[n][4]:=Sum([1,2,5,7,8],p->M[n-1][p]);
M[n][5]:=Sum([1,2,3,4,6,7,8,9],p->M[n-1][p]);
M[n][6]:=Sum([2,3,5,8,9],p->M[n-1][p]);
M[n][7]:=Sum([4,5,8,10],p->M[n-1][p]);
M[n][8]:=Sum([4,5,6,7,9,10],p->M[n-1][p]);
M[n][9]:=Sum([5,6,8,10],p->M[n-1][p]);
M[n][10]:=Sum([7,8,9],p->M[n-1][p]);
od;
In a previous version of the question, there was a condition on the 10-digits being unique. This is a much harder problem, and amounts to counting the number of hamilton paths in various $10$-vertex graphs. Note that finding a hamilton path is an NP-complete problem (ref.). Nevertheless, this enumeration should be achievable using a backtracking algorithm (judging from the above numbers). There's probably not a much better way than this; see these Stackoverflow questions here and here.
Best Answer
Your answer to the first question is far too high: you have five main choices, two side choices and four drink choices, making $5 \times 2 \times 4$ overall possible combinations, since you want one of each.
On the second, the answer is yes. Note that if you want no matches week-to-week then you will have to alternate sides each week, but you have enough flexibility on mains and drinks to do this. It is not too difficult to make a list of $40$ combinations to prove this. Here is one possibility:
HFS CAT NFJ WAW SFS HAT CFJ NAW WFS SAT HFJ CAW NFS WAT SFJ HAW CFS NAT WFJ SAW HFT CAS NFW WAJ SFT HAS CFW NAJ WFT SAS HFW CAJ NFT WAS SFW HAJ CFT NAS WFW SAJ