Hints:
The failure of the batteries is a Poisson process
Having two batteries on at the same time doubles the intensity of the process
You can use the $n$ batteries up to the point at which there have been $n-1$ failures
The problems are not all about "max." The first one is about "min." The system fails as soon as a component fails. So its lifetime is the minimum of the lifetimes of the components.
Let $X_1,X_2, \dots,X_n$ be identically distributed independent random variables, with continuous distribution.
Let $Y$ be the minimum of the $X_i$, and let $Z$ be the maximum of the $X_i$. We want to find the distribution of $Y$ and of $Z$.
For any $i$, let $F$ be the cumulative distribution function of $X_i$ (they are all the same.)
The minimum: We have $Y\gt y$ if and only if all the $X_i$ are $\gt Y$. For any $i$, the probability that $X_i \gt y$ is $1-F(y)$. So the probability they are all $\gt y$ is $(1-F(y))^n$. Thus the probability that $Y\le y$ is equal to $1-(1-F(y))^n.$
If we can calculate $F$, we know the cdf of $Y$, and therefore, if we need it, the density function of $Y$. Once we have these, most common problems about $Y$ are reasonably straightforward.
The maximum: This is simpler. The maximum $Z$ is $\le z$ if and only if all the $X_i$ are $\le z$. the probability of this is $(F(z))^n$. So the cumulative distribution function of $Z$ is $(F(z))^n$.
All three of your problems can be solved using the facts about the distribution of $Y$ and $Z$ just derived.
One does not need to remember the formulas: in any particular case, we can repeat the reasoning that led to the formulas.
If you need the calculation for any specific one of the three problems, it can be added.
Best Answer
The distribution given by $f(x)$ is $\exp(λ=1/200)$. So,
Now, for your second question, denote with $X_1$ the minimum, with $X_2$ the middle and with $X_3$ the maximum lifetime of the three components. As is already mentioned you need to calculate $E[X_3]$. We will use that the minimum of $n$ iid $\exp(λ)$ random variables has again the exponential distribution with parameter $nλ$, see here. Write $$E[X_3]=E[X_1+(X_2-X_1)+(X_3-X_2)]=E[X_1]+E[X_2-X_1]+E[X_3-X_2]$$ by linearity of expectation. Now,
$X_1$ is the minimum of $3$ iid $\exp(λ)$ hence $X_1\sim\exp(3λ)$ and
$X_2-X_1$ is distributed as (by the memoryless property of the exponential) as the minimum of $2$ iid $\exp(λ)$, hence $X_2-X_1\sim \exp(2λ)$. Why is that? Because after $X_1$ is realized, you have two more components running. By the memoryless property of the exponential distribution, the remaining lifetime of each of the components is again exponentially distributed. So, $X_2-X_1$ is just the time that the minimum of $2$ iid exponentials will be realised.
By the same argument $X_3-X_2$ is distributed as $X_3-X_2\sim\exp(λ)$.
So, $$E[X_3]=\frac1{3λ}+\frac1{2λ}+\frac1{λ}=\frac{200}3+\frac{200}2+200$$