What can we say about the set of real numbers lying between $\sqrt{5}$ and $\sqrt{8}$ in the Markov spectrum?
$$ \left\{ \xi \in \mathbb{R} : \frac{1}{\sqrt{8}n^2}< \left| \xi - \frac{m}{n}\right|< \frac{1}{\sqrt{5}n^2} \text{ for infinitely many }m,n\in \mathbb{Z}\right\}\subseteq \mathbb{R} $$ What is the Hausdorff dimension of this set? put another way we're looking for real numbers such that
$$ \frac{1}{\sqrt{8}}< \liminf_{n \to \infty} n^2 \left| \xi - \frac{m}{n}\right|< \frac{1}{\sqrt{5}} $$
Perhaps this can be solved with continued fractions.
Related
Continued Fractions
For continued fractions, what is true is that for any irrational $\xi$, the $n^\text{th}$ convergent, $\frac{p_n}{q_n}$ satisfies $$ \frac1{(a_{n+1}+2)\,q_n^2}\lt\left|\,\xi-\frac{p_n}{q_n}\,\right|\lt\frac1{a_{n+1}q_n^2}\tag1 $$ where $a_{n+1}$ is the $n+1^\text{st}$ term in the continued fraction.
For example, $\pi=(3,7,15,1,292,\dots)$ and $(3,7,15,1)=\frac{355}{113}$. Then $$ \begin{array}{c} \frac1{294\cdot113^2}\lt\left|\,\pi-\frac{355}{113}\,\right|\lt\frac1{292\cdot113^2}\\[6pt] 0.000000266376\lt0.000000266764\lt0.000000268201 \end{array}\tag2 $$
If we have $$ \liminf_{q\to\infty}q^2\left|\,\xi-\frac pq\,\right|\ge\frac1{\sqrt8}\tag3 $$ then, for all but finitely many $n$, the terms of the continued fraction of $\xi$ must be $1$ or $2$. That is, if infinitely many are $3$ or greater, the $\liminf$ would be at most $\frac13$.
Suppose the $n-2^\text{nd}$ and $n-1^\text{st}$ convergents for a continued fraction are $\frac{p_{n-2}}{q_{n-2}}$ and $\frac{p_{n-1}}{q_{n-1}}$. Numbers with these convergents can be written as $\frac{p_{n-2}+r_np_{n-1}}{q_{n-2}+r_nq_{n-1}}$ where $r_n\ge1$. The next term in the continued fraction is $a_n=\lfloor r_n\rfloor$. The fraction of real numbers whose continued fraction start the same but whose next term is $a_n$ is $$ \begin{align} \frac{\frac{p_{n-2}+(a_n+1)p_{n-1}}{q_{n-2}+(a_n+1)q_{n-1}}-\frac{p_{n-2}+a_np_{n-1}}{q_{n-2}+a_nq_{n-1}}}{\frac{p_{n-1}}{q_{n-1}}-\frac{p_{n-2}+p_{n-1}}{q_{n-2}+q_{n-1}}} &=\frac{q_{n-1}(q_{n-2}+q_{n-1})}{(q_{n-2}+a_nq_{n-1})(q_{n-2}+(a_n+1)q_{n-1})}\\ &=\frac{1+x}{(a_n+x)(a_n+1+x)}\quad\text{where }x=\frac{q_{n-2}}{q_{n-1}}\in[0,1]\\[6pt] &\in\left\{\begin{array}{} \left[\frac13,\frac12\right]&\text{if }a_n=1\\ \left[\frac16,3-2\sqrt2\right]&\text{if }a_n=2\\ \left[\frac1{a_n(a_n+1)},\frac2{(a_n+1)(a_n+2)}\right]&\text{if }a_n\ge3 \end{array}\right.\tag4 \end{align} $$ Note that we approach $3-2\sqrt2$ in the case $a_n=2$ when $\frac{q_{n-2}}{q_{n-1}}\to\sqrt2-1$.
For example, consider the continued fractions which start $(0,2,3,\dots)$. These are the reals in the interval $\left[\frac37,\frac49\right]$. The continued fractions that start $(0,2,3,\color{#C00}{2},\dots)$ are in the interval $\left[\frac{10}{23},\frac7{16}\right]$. The proportion of the numbers in the first range that are in the second range is $$ \frac{\frac7{16}-\frac{10}{23}}{\frac49-\frac37}=\frac{63}{368}\tag5 $$ Formula $(4)$ says that this proportion, for $a_n=\color{#C00}{2}$, should be in the range $\left[\frac16,3-2\sqrt2\right]$, and indeed, it is.
Here is an image showing the real numbers whose continued fractions start with $1$ or $2$:
Fractal Dimension
Suppose that we have a self-similar replication with two different ratios, $0\lt q\le p\lt1$, where $p+q\le1$.
Let's count the number of pieces whose size is $\frac1{2^n}\sim p^aq^b$, that is, $$ a\log(p)+b\log(q)\sim n\log\left(\frac12\right)\tag6 $$ For each $a$ and $b$, we have $\binom{a+b}{a}$ ways to arrange the factors of $p$ and $q$ to get the different segments. Thus, we want to compute $$ \sum_{k=0}^{\frac{n_0}{1-\gamma}}\binom{n_0+k\gamma}{k}\tag7 $$ where $$ n_0=\frac{n\log\left(\frac12\right)}{\log(q)}\quad\text{and}\quad1-\gamma=\frac{\log(p)}{\log(q)}\tag8 $$ The ratio of consecutive terms of $(7)$ is $$ \frac{\binom{m+\gamma}{k+1}}{\binom{m}{k}}\sim\frac{\left(1-\frac km\right)^{1-\gamma}}{\frac km}\tag9 $$ where $m=n_0+k\gamma$. Therefore, to find the greatest term in $(7)$, we need to compute $$ (1-x)^{1-\gamma}=x\tag{10} $$ where $x=\frac km$. Since $x=\frac{k}{n_0+k\gamma}$, we have $k=\frac{n_0\,x}{1-x\gamma}$ and $m=n_0+k\gamma=n_0+\frac{n_0\,x\gamma}{1-x\gamma}$. That is, $$ m=\frac{n_0}{1-x\gamma}\tag{11} $$ Using $(8)$, we get $1-x\gamma=\frac{x\log(p)+(1-x)\log(q)}{\log(q)}$ and $$ m=\frac{n\log\left(\frac12\right)}{x\log(p)+(1-x)\log(q)}\tag{12} $$ Let's compute how the terms of $(7)$ behave near the maxiumum, i.e. $k=mx$: $$ \begin{align} \frac{\left(1-\frac{k+1}{m+\gamma}\right)^{1-\gamma}}{\frac{k+1}{m+\gamma}}\frac{\frac km}{\left(1-\frac km\right)^{1-\gamma}} &=\frac{\left(1-\frac{m-k\gamma}{m(m-k)}\right)^{1-\gamma}}{1+\frac{m-k\gamma}{km}}\\ &\sim1-\frac{(m-k\gamma)^2}{mk(m-k)}\\ &=1-\frac{(1-x\gamma)^2}{mx(1-x)}\tag{13} \end{align} $$ This means that for $k=mx$, $$ \binom{m+j\gamma}{k+j}\sim\binom{m}{k}e^{-\frac{(1-x\gamma)^2}{2mx(1-x)}\,j^2}\tag{14} $$ Summing the right side of $(14)$ yields $$ \sum_{j=-\infty}^\infty e^{-\frac{(1-x\gamma)^2}{2mx(1-x)}\,j^2}\sim\frac{\sqrt{2\pi mx(1-x)}}{1-x\gamma}\tag{15} $$ Thus, using Stirling's approximation, $(14)$, and $(15)$, we can approximate the sum in $(7)$ as $$ \frac1{(1-x\gamma)\left(x^x(1-x)^{1-x}\right)^m}\tag{16} $$ The fractal dimension of this self-similarity is the limit of the log of the number of pieces divided by the negative of the log of the size of the pieces. Using $(12)$ and $(16)$, we get $$ \begin{align} d &=\lim_{n\to\infty}\frac{m(x\log(x)+(1-x)\log(1-x))+\log(1-x\gamma)}{n\log\left(\frac12\right)}\\[6pt] &=\frac{x\log(x)+(1-x)\log(1-x)}{x\log(p)+(1-x)\log(q)}\tag{17} \end{align} $$
For example, let's apply $(17)$ to the Cantor set, which has $p=q=\frac13$. Equation $(8)$ gives $\gamma=0$. Equation $(9)$ gives $x=\frac12$. Finally, equation $(17)$ gives $d=\frac{\log\left(\frac12\right)}{\log\left(\frac13\right)}=\frac{\log(2)}{\log(3)}$, as expected.
Another example is the asymmetric Cantor set, which has $p=\frac12$ and $q=\frac14$. Equation $(8)$ gives $\gamma=\frac12$. Equation $(9)$ gives $x=\frac1\phi$. Note that $1-\frac1\phi=\frac1{\phi^2}$. Thus, equation $(17)$ gives $d=\frac{\frac1\phi\log\left(\frac1\phi\right)+\left(1-\frac1\phi\right)\log\left(\frac1{\phi^2}\right)}{\frac1\phi\log\left(\frac12\right)+\left(1-\frac1\phi\right)\log\left(\frac14\right)}=\frac{\log(\phi)}{\log(2)}$, which agrees with this Wikipedia article.
The Fractal Dimension of the Continued Fractions
The dimension we are looking for would be between the dimensions for self-similarities with ratios $\frac13$ and $\frac16$ and ratios $\frac12$ and $3-2\sqrt2$.
Setting $p=\frac13$ and $q=\frac16$ in $(17)$ gives $$ d=0.4895363211996494887\tag{18} $$ Setting $p=\frac12$ and $q=3-2\sqrt2$ in $(17)$ gives $$ d=0.6065975116111343606\tag{19} $$