probability infected by two disease with assumption Assumption:

A people contacts with $k$ neighbours. $m$ of these neighbours have infected with disease A and $n$ of them have infected with disease B. The rest which keeps healthy is $k-m-n$.
People infect disease A from neighbours with disease A in probability $\lambda_A$.
People infect disease B from neighbours with disease B in probability $\lambda_B$.
I:
A people can have disease A and disease B at the same time(e.g. flu$_A$, Pulmonary tuberculosis$_B$). So, there are four cases:
Not infected with disease A and disease B. ($\overline{A}\overline{B}$): $(1-\lambda_A)^m (1-\lambda_B)^n$
Infected with disease A and disease B. ($AB$): $(1-(1-\lambda_A)^m) (1-(1-\lambda_B)^n)=1-(1-\lambda_A)^m-(1-\lambda_B)^n+(1-\lambda_A)^m (1-\lambda_B)^n$
Infected with disease A but not infected with disease B. ($A\overline{B}$): $(1-(1-\lambda_A)^m) (1-\lambda_B)^n=(1-\lambda_B)^n-(1-\lambda_A)^m (1-\lambda_B)^n$
Infected with disease B but not infected with disease A. ($\overline{A}B$): $(1-\lambda_A)^m (1-(1-\lambda_B)^n)=(1-\lambda_A)^m-(1-\lambda_A)^m (1-\lambda_B)^n$
So, $P(\overline{A}\overline{B})+P(AB)+P(A\overline{B})+P(\overline{A}B)=1$.
$P(A)+P(B)-P(AB)+P(\text{Not Infected})=P(AB)+P(A\overline{B})+P(AB)+P(\overline{A}B)-p(AB)+P(\overline{A}\overline{B})=1$.
Are these probabilities right in these cases?
II:
A people can only one of two diseases at the same time (like flu$_A$, flu$_B$). So:
Not infected with disease A and disease B. ($\overline{A}\overline{B}$): $(1-\lambda_A)^m (1-\lambda_B)^n$  Right??
Infected with disease A. ($P(A|\overline{B})=\frac{P(A,\overline{B})}{P(\overline{B})}$ or just $P(A)$) ?? And what is the detail?
Infected with disease B. ($P(B|\overline{A})=\frac{P(\overline{A},B)}{P(B)}$ or just $P(B)$) ?? And what is the detail?
EDIT: $P(A)+P(B)+P(\text{Not Infected}) = 1$?
Do I need to supply more information about my problem? Please comment below. Thanks for your time.
 A: Your part I is correct. (For being infected with both disease $A$ and $B$ I prefer the notation $P(A\cap B)$ or $P(A,B)$ to $P(AB)$, however I will use your notation).
For part II:
The probability of being infected with disease $A$ has notation $P(A)$; $P(A|\overline{B})$ means the probability of being infected with disease $A$ given that you are not infected with disease $B$.
Just like the previous part, the probability of not being infected by disease $A$ is $P(\overline{A})=(1-\lambda_A)^m$. Therefore the complementary event has probability $P(A)=1-(1-\lambda_A)^m$. With the same arguments we see that $P(\overline{B})=(1-\lambda_B)^n$ and $P(B)=1-(1-\lambda_B)^n$. Your EDIT is correct, as you can only be infected by either disease $A$, $B$ or none.
Using that fact, we obtain $P(\overline{AB})=1-P(A)-P(B)=(1-\lambda_A)^m-(1-\lambda_B)^n-1$.
The same could be concluded from the total law of probability: $P(\overline{A})=P(\overline{A}B)+P(\overline{AB})=P(\overline{A}|B)P(B)+P(\overline{AB})=P(B)+P(\overline{AB})$, so $P(\overline{AB})=P(\overline{A})-P(B)=(1-\lambda_A)^m-(1-\lambda_B)^n-1$, because if you are infected by disease $B$, then automatically you cannot be infected by $A$, so $P(\overline{A}|B)=1$.
Remarks:
$P(\overline{AB})\neq (1-\lambda_A)^m (1-\lambda_B)^n$. You cannot reason towards that conclusion, because you can't reason them to be indepedent (for $\lambda_A,\lambda_B>0)$.
You could also try to find all of the conditional probabilities. For example, $P(A|B)=P(B|A)=0$, because given being infected with disease $A$, you cannot be infected with disease $B$. For the other ones, use $P(\overline{A}|\overline{B})=\frac{P(\overline{AB})}{P(\overline{B})}$ and $P(A|\overline{B})=\frac{P(A\overline{B})}{P(\overline{B})}=\frac{P(\overline{B}|A)P(A)}{P(\overline{B})}=\frac{P(A)}{P(\overline{B})}$.
A: Too long for a comment. 
II
I think this is a hard question. 
“- What is a probability to meet a crocodile at your street?”
“- One over two. I either meet him, or do not meet.”

As it sometimes happens for probability problems, I think that the problem formulation is incomplete. In particular, it is not clear how to choose whether the person will be infected with decease $A$ or decease $B$ if both cases are possible. The problem formulation can be completed by providing an explicit and exact model. But this approach raises an other problem, because
different models may lead to different values of probability. A famous example is “the Bertrand paradox, [which] is a problem within the classical interpretation of probability theory. Joseph Bertrand introduced it in his work ‘Calcul des probabilités’ (1889) as an example to show that probabilities may not be well defined if the mechanism or method that produces the random variable is not clearly defined”.
I think the following model is natural. A person consecutively visits all its sick neighbors. For instance, because he is Santa delivering Christmas gifts and all his neighbors were good during the last year. (Don’t be upset because of Santa. Even if he’ll be infected, he’ll recover soon. He is strong. HO HO HO!)  
In order to be infected by a decease it is relevant only an order $\sigma$ in which Santa visits his sick neighbors. Since he loves each of them equally, Santa chooses as $\sigma$ a permutation of a symmetric group $S_{m+n}$ with an equal probability $ 1/|S_{m+n}|=1/(m+n)!$ for each permutation $\sigma$. 
The probability $1-P(A)-P(B)$ not to be infected after all visits does not depend on $\sigma$ and equals $(1-\lambda_A)^m(1-\lambda_B)^n$. But for a given permutation $\sigma=(\sigma_i)\in S_{m+n}$ of neighbors the probability that $r$-th neighbor was (the first) who infected Santa is $\lambda_{\sigma,r}\prod_{i=1}^{r-1} (1-\lambda_{\sigma,i})$, where $\lambda_{\sigma,j}=\lambda_A$, if $\sigma_i$-th neighbor is infected by decease $A$ and $\lambda_{\sigma,j}=\lambda_B$, if $\sigma_i$-th neighbor is infected by decease $B$. Thus the probability that Santa will be infected by disease $A$ is $$P(\sigma,A)=\sum_{\sigma_r=A} \lambda_{\sigma,r} \prod_{i=1}^{r-1} (1-\lambda_{\sigma,i}).$$ The total probability $P(A)$ for Santa to be infected with desease $A$ is 
$$\frac 1{(m+n)!}\sum_{\sigma \in S_{m+n} } P(\sigma,A)=\frac 1{(m+n)!}\sum_{\sigma \in S_{m+n}} \sum_{\sigma_r=A} \lambda_A \prod_{i=1}^{r-1} (1-\lambda_{\sigma,i})$$
The latter sum should be equal to something like 
$$\frac 1{(m+n)!} \sum_{r=1}^{m+n} \lambda_A \sum_{j=0}^{r-1} (1-\lambda_A)^j(1-\lambda_B)^{r-1-j}
m{r-1\choose j}{m-1\choose j}j!\times$$ $${n\choose r-1-j}(r-1-j)!(m+n-r)!$$
which looks hard to calculate even when $\lambda_A=\lambda_B$.
Another approach to calculate $P(A)=P(m,n,A)$ is to use a recurrent formula for each $m,n\ge 1$ 
$$P(m,n,A)=\frac{1}{m+n}(m(\lambda_A+(1-\lambda_A) P(m-1,n,A))+n(1-\lambda_B)P(m,n-1,A)) )$$
with boundary conditions $P(m,0,A)=1-(1-\lambda_A)^m$ and $P(0,n,A)=0$. 
By the way, the formula is similar to the recurrent formula for $E(m,n)$ from my question on extrasensory perception strategy, leading to very non-trivial estimations even when $m=n$ and the probabities are equal to $1/2$. 
I can provide a few results for our case. Namely, when $n=m$  and $\lambda_A=\lambda_B=\lambda$ then by symmetry $P(A)=P(B)=\frac 12(1-(1-\lambda)^{m+n})$.
Also for a particular case $\lambda_A=\lambda_B=1/2$ and $m=1$ by an easy induction we can show that $P(1,n,A)=\frac 1{n+1}\left(1-\frac 1{2^{n+1}}\right)$ for each $n\ge 0$. But already for $P(2,n,A)$ I expect a more complicated formula. 
A: Your answers to the first part are spot on.
In the second part, as a person can be infected with only one disease at a time, that means, the probability she is infected with disease A = $1$- probability of not getting infected with disease A = $1-(1-\lambda_A)^m$.
Similar is the case for the second disease.
