In fact, it is always true that for two random variables $X$ and $Y$ (independent or not), $\text{Var}(\max(X,Y)) + \text{Var}(\min(X,Y)) \le \text{Var}(X) + \text{Var}(Y)$. This gives you the upper bound. I don't see an easy way to get the lower bound.
The Poisson with paramteter $\lambda$ has mean $\lambda$. So when we are approximating a binomial with parameters $n$, $p$ (and therefore mean $np$) by a Poisson, the appropriate parameter $\lambda$ is the mean $np$ of the binomial. But the above can mainly be thought of as a mnemonic, a device to remember the right answer. So we go into more detail.
The following is an informal calculation that can be turned into a formal limit argument. If $X$ is the binomial, then
$$\Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}.\tag{1}$$
Use the abbreviation $\lambda$ for $np$. Then $p=\frac{\lambda}{n}$. So we have
$1-p=1-\frac{\lambda}{n}$.
We can then rewrite (1) as
$$\Pr(X=k)=\frac{1}{k!}(n)(n-1)\cdots(n-k+1)\frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^n \left(1-\frac{\lambda}{n}\right)^{-k}.\tag{2}$$
Note that we have done nothing so far, except for the introduction of $\lambda$ as an abbreviation for $np$.
Now imagine $n$ large, and $p$ small, so that $np=\lambda$ stays constant. Let $k$ be fixed. We look at the various terms in Formula (2).
For fixed $k$, if $n$ is large then $(n)(n-1)\cdots(n-k+1)\frac{1}{n^k}\approx 1$ and $\left(1-\frac{\lambda}{n}\right)^{-k}\approx 1$. Also, for $n$ large, $\left(1-\frac{\lambda}{n}\right)^n \approx e^{-\lambda}$.
It follows that the right-hand side of (2) is approximately
$$\frac{1}{k!} \lambda^ke^{-\lambda}.$$
This is precisely the probability that a Poisson with parameter $\lambda$ takes on the value $k$.
In your example, the mean number of rotten apples in a batch of $100$ is $3$, and $n=100$ is reasonably large, so one can expect the Poisson with parameter $3$ to provide a good fit, at least for reasonable $k$.
Remark: We gave a fair bit of detail justifying why in the case $n$ large, $p$ small, and $np=\lambda$ moderate, we can approximate binomial distribution probabilities by Poisson distribution probabilities. But the main answer to your question is that the appropriate $\lambda$ is always the mean of your binomial. If a superbushel holds $150$ apples, the appropriate $\lambda$ is $4.5$.
Best Answer
The mean per page is $2.5$, so the mean per $30$ pages is $(30)(2.5)$.
This really has nothing much to do with the Poisson. Let $X_1$ be the number of typos on the first page, $X_2$ the number of typos on the second, and so on up to $30$. The total number $T$ of typos is given by $$T=X_1+X_2+\cdots+X_{30}.$$ The mean of a sum is always the sum of the means ("linearity of expectation"). It follows that $E(T)=30E(X_1)=75$.
Where the Poisson comes in is not for the computation of the mean. But it is important for the distribution of $T$. Since the sum of independent Poissons is Poisson, $T$ has Poisson distribution.