I will say that this solution seems a bit advanced for an introductory course, but I cannot see any other reasonable way to interpret these questions.
Perhaps someone else can provide an answer for something, I am not seeing.
In either case it can serve as a very short crash course introduction to Poisson processes.
We know that the number of imperfections on a $1$cm wire, call it $\mathsf{N}_1$, is Poisson with parameter $28$, i.e. $\mathsf{N}_1\sim \mathrm{Poisson}(\lambda)$ with $\lambda=28$.
Let's say that you want to know the number of imperfection on a wire that is $t$ cm long for some $t\in[0, \infty)$, call it $\mathsf{N}_t$.
Intuitively, you would expect this to be Poisson as well, but with a scaled parameter, i.e. $\mathsf{N}_t\sim \mathrm{Poisson}(t\cdot\lambda)$.
Now you're not really interested in the number of faults, but rather the distances between them.
Let's call these $\mathsf{X}_1, \mathsf{X}_2, \ldots$ and assume that they are independent.
Now note the important relationship that $\mathsf{N}_t=k$ if and only if $\mathsf{X}_1+\mathsf{X}_2+\ldots+\mathsf{X}_k\le t$, but $\mathsf{X}_1+\mathsf{X}_2+\ldots+\mathsf{X}_k+\mathsf{X}_{k+1}>t$ (making a drawing might be helpful).
Let's try to find the distribution of these $\mathsf{X}_i$'s by induction in $i$.
First, note that $\mathsf{X}_1\le t$ if and only if $\mathsf{N}_t\ge1$, i.e. if and only if there is at least one fault in the first $t$ centimeters.
As such, we have that
$$
\mathbb{P}(\mathsf{X}_1\le t)
=\mathbb{P}(\mathsf{N}_t\ge 1)
=1- \mathbb{P}(\mathsf{N}_t=0)
=1- \mathrm{e}^{-t\cdot\lambda}.
$$
This is the CDF of the $\mathrm{Exponential}(\lambda)$-distribution!
This means that $\mathsf{X}_1\sim \mathrm{Exponential}$, and I claim this to be true for all the $\mathsf{X}_i$'s.
As such, assume that it is true for $\mathsf{X}_1,\ldots,\mathsf{X}_k$, and let's prove it for $\mathsf{X}_{k+1}$.
For notation set $\mathsf{S}_k=\sum_{i=1}^{k}\mathsf{X}_i$.
Using the important relationship from before, we find that
$$
\mathbb{P}(\mathsf{N}_t=k)
=\mathbb{P}(\mathsf{S}_k\le t, \mathsf{S}_k+\mathsf{X}_{k+1}>t)
=\mathbb{P}(\mathsf{S}_k\le t, \mathsf{X}_{k+1}>t-\mathsf{S}_k).
$$
Using the law of total probability and the independence of $\mathsf{X}_{k+1}$ and $\mathsf{S}_k$, we may write the right hand side as
$$
\mathbb{P}(\mathsf{S}_k\le t, \mathsf{X}_{k+1}>t-\mathsf{S}_k)
=\int_{0}^{t}\mathbb{P}(\mathsf{X}_{k+1}>t-s)f_{\mathsf{S}_k}(s)\mathrm{d}s,
$$
and by our induction hypothesis, $\mathsf{S}_k\sim\Gamma(k, \lambda)$, i.e. $f_{\mathsf{S}_k}(s)=\frac{\lambda^k}{(k-1)!}s^{k-1}\mathrm{e}^{-\lambda s}$.
Plugging in also our ansatz that $\mathsf{X}_{k+1}\sim \mathrm{Exponential}(\lambda)$, and hence $\mathbb{P}(\mathsf{X}_{k+1}>t-s)=\mathrm{e}^{-\lambda(t-s)}$, this becomes
\begin{align*}
\int_{0}^{t}\mathbb{P}(\mathsf{X}_{k+1}>t-s)f_{\mathsf{S}_k}(s)\mathrm{d}s
&=\frac{\lambda^k}{(k-1)!}\int_{0}^{t}\mathrm{e}^{-\lambda(t-s)}s^{k-1}\mathrm{e}^{-\lambda s}\mathrm{d}s \\
&=\frac{\lambda^k \mathrm{e}^{-\lambda t}}{(k-1)!}\int_{0}^{t}s^{k-1}\mathrm{d}s \\
&=\frac{(t\lambda)^k \mathrm{e}^{-\lambda t}}{k!},
\end{align*}
but this is exactly $\mathbb{P}(\mathsf{N}_t=k)$ as desired, and hence $\mathsf{X}_{k+1}\sim \mathrm{Exponential}(\lambda)$ as well.
But this is where your questions confuse me.
Since the gaps between faults are continuous (exponential) the probability that they are exactly $0.025$ or $0.043$ are $0$, making the first two questions rather boring.
To get a more fulfilling answer, they really should be inequalities (e.g. the probability of a distance of less than $0.025$cm).
However for the final question, the average will converge to the mean of the distance, i.e. the mean of $\mathrm{Exponential}(\lambda)$, so $\frac{1}{\lambda}=\frac{1}{28}$.
I hope that this was more helpful than confusing.
