Divide a square using horizontal and vertical lines into 16 equal smaller squares as follow:
A counter is placed at random on one of these squares and is then moved $n$ times. At each of these moves it can be transferred to any neighbouring square, in any of the 8 directions with equal probability.
let $c_n$ be the probability that a particular corner site is occupied after $n$ such independent moves, and let the corresponding probabilities for an intermediate site at the side of the board and for a site in the middle of the board be $s_n$ and $m_n$ respectively. Show that
$4c_n+8s_n+4m_n=1$, $n=0,1,2,...,$. and that $c_n=\frac{2}{5}s_{n-1}+\frac{1}{8}m_{n-1}$, $n=1,2,...$. Find two other relations for $s_n$ and $m_n$ in terms of $c_{n-1}, s_{n-1}$, and $m_{n-1}$ and hence find $c_n, s_n$ and $m_n$. (Oxford 1974M).
I think $c_n=\frac{2}{5}s_{n-1}+\frac{1}{8}m_{n-1}$ should be $c_n=\frac{1}{5}s_{n-1}+\frac{1}{8}m_{n-1}$
And $s_n=s_{n-1}\frac{2}{5}+c_{n-1}\frac{2}{3}+m_{n-1}\frac{1}{2}$
And $m_n=m_{n-1}\frac{3}{8}+s_{n-1}\frac{2}{5}+c_{n-1}\frac{1}{3}$.
Also, $c_0=\frac{1}{4}, s_0=\frac{1}{2}, m_n=\frac{1}{4}$.
How do we solve for $c_n,s_n,m_n$?
Note that $c_n,s_n,m_n$ are the respective probabilities, starting from a random cell, of ending, after $n$ moves, in a particular corner cell, side cell, or middle cell.
Let $C_n,S_n,M_n$ be the the respective probabilities, starting from a random cell, of ending, after $n$ moves, in some corner cell, side cell, or middle cell (where we don't care which one).
It's automatic that $$C_n+S_n+M_n=1$$ and clearly we have $$ \begin{cases} C_n=4c_n\\[4pt] S_n=8s_n\\[4pt] M_n=4m_n\\ \end{cases} $$ hence $$4c_n+8s_n+4m_n=1$$ which resolves the first part of the problem.
Note that in your attempt, the recursions you came up with are really for $C_n,S_n,M_n$, rather than for $c_n,s_n,m_n$.
Thus, for $C_n,S_n,M_n$ we get the initial conditions $$C_0=\frac{1}{4},\;\;\;S_0=\frac{1}{2},\;\;\;M_0=\frac{1}{4}$$ together with the recurrence relations $$ \begin{cases} C_n = {\large{\frac{1}{5}}}S_{n-1} + {\large{\frac{1}{8}}}M_{n-1} \\[4pt] S_n = {\large{\frac{2}{5}}}S_{n-1} + {\large{\frac{2}{3}}}C_{n-1} + {\large{\frac{1}{2}}}M_{n-1}\\[4pt] M_n = {\large{\frac{3}{8}}}M_{n-1} + {\large{\frac{2}{5}}}S_{n-1} + {\large{\frac{1}{3}}}C_{n-1} \\ \end{cases} $$ for $n\ge 1$.
For $c_n,s_n,m_n$, the corresponding results are $$c_0=\frac{1}{16},\;\;\;s_0=\frac{1}{16},\;\;\;m_0=\frac{1}{16}$$ together with the recurrence relations $$ \begin{cases} c_n = {\large{\frac{2}{5}}}s_{n-1} + {\large{\frac{1}{8}}}m_{n-1} \\[4pt] s_n = {\large{\frac{2}{5}}}s_{n-1} + {\large{\frac{1}{3}}}c_{n-1} + {\large{\frac{1}{4}}}m_{n-1}\\[4pt] m_n = {\large{\frac{3}{8}}}m_{n-1} + {\large{\frac{4}{5}}}s_{n-1} + {\large{\frac{1}{3}}}c_{n-1} \\ \end{cases} $$ for $n\ge 1$.
For each nonnegative integer $k$, let $v_k$ denote the vector $\langle{c_k,s_k,m_k}\rangle$.
Then in matrix form, for $n\ge 1$, we have $v_n=Tv_{n-1}$ where $$ T= {\large{ \pmatrix { 0 & \frac{2}{5} & \frac{1}{8} \cr \frac{1}{3} & \frac{2}{5} & \frac{1}{4} \cr \frac{1}{3} & \frac{4}{5} & \frac{3}{8} \cr } }} $$ It follows that for all positive integers $n$, we have $v_n=T^nv_0$.
Noting that $T$ has minimal polynomial $$t^3-{\small{\frac{31}{40}}}t^2-{\small{\frac{9}{40}}}t$$ we get $$T^3={\small{\frac{31}{40}}}T^2+{\small{\frac{9}{40}}}T$$ hence, for all integers $n\ge 3$, \begin{align*} &T^n={\small{\frac{31}{40}}}T^{n-1}+{\small{\frac{9}{40}}}T^{n-2}\\[4pt] \implies\;&T^nv_0={\small{\frac{31}{40}}}T^{n-1}v_0+{\small{\frac{9}{40}}}T^{n-2}v_0\\[4pt] \implies\;&v_n={\small{\frac{31}{40}}}v_{n-1}+{\small{\frac{9}{40}}}v_{n-2}\\[4pt] \end{align*} which yields the recurrence relations \begin{align*} c_n&= \begin{cases} {\large{\frac{1}{16}}}&&\;\;\;\;\text{if}\;\;n=0\\[4pt] {\large{\frac{21}{640}}}&&\;\;\;\;\text{if}\;\;n=1\\[4pt] {\large{\frac{931}{25600}}}&&\;\;\;\;\text{if}\;\;n=2\\[4pt] {\large{\frac{31}{40}}}c_{n-1}+{\large{\frac{9}{40}}}c_{n-2}&&\;\;\;\;\text{if}\;\;n\ge 3\\ \end{cases} \\[14pt] s_n&= \begin{cases} {\large{\frac{1}{16}}}&&\;\;\;\;\text{if}\;\;n=0\\[4pt] {\large{\frac{59}{960}}}&&\;\;\;\;\text{if}\;\;n=1\\[4pt] {\large{\frac{2269}{38400}}}&&\;\;\;\;\text{if}\;\;n=2\\[4pt] {\large{\frac{31}{40}}}s_{n-1}+{\large{\frac{9}{40}}}s_{n-2}&&\;\;\;\;\text{if}\;\;n\ge 3\\ \end{cases} \\[14pt] m_n&= \begin{cases} {\large{\frac{1}{16}}}&&\text{if}\;\;n=0\\[4pt] {\large{\frac{181}{1920}}}&&\text{if}\;\;n=1\\[4pt] {\large{\frac{7331}{76800}}}&&\text{if}\;\;n=2\\[4pt] {\large{\frac{31}{40}}}m_{n-1}+{\large{\frac{9}{40}}}m_{n-2}&&\text{if}\;\;n\ge 3\\ \end{cases} \\[4pt] \end{align*} Finally, using standard linear-recurrence-solving techniques, together with the initial values $$c_0=\frac{1}{16},\;\;\;s_0=\frac{1}{16},\;\;\;m_0=\frac{1}{16}$$ we get the formulas \begin{align*} c_n&=\frac{1}{28}+\frac{13}{1008}\cdot\left(-\frac{9}{40}\right)^n\\[4pt] s_n&=\frac{5}{84}-\frac{13}{1512}\cdot\left(-\frac{9}{40}\right)^n\\[4pt] m_n&=\frac{2}{21}+\frac{13}{3024}\cdot\left(-\frac{9}{40}\right)^n\\[4pt] \end{align*} valid for all positive integers $n$.