Probability of choosing bat 
I got the option A and B by manually checking but after that got stuck .
 A: Having used B1 in the first match he is not allowed to use it in the second match, justifying $P(2)=0$.  As he has not used it in the second match he has three to choose from for the third and B1 is one of them so $P(3)=\frac 13$.  Now for the fourth match he has $\frac 23$ chance of not using bat B1 in the third match.  For that chance he has $\frac 13$ chance of using it for the fourth, so $P(4)=\frac 23 \cdot \frac 13=\frac 29$.  Generally, if he has chance $p$ of using B1 in a match, he has $\frac 13(1-p)$ chance of using it in the next match because if he does use B1 in the match he won't use it the next time and if he does not use it he has $\frac 13$ chance of using it next time.  Thus we get $P(5)=\frac 13(1-\frac 29)=\frac 7{27}, P(6)=\frac 13(1-\frac 7{27})=\frac {20}{81}, P(7)=\frac 13(1-\frac {20}{81})=\frac {61}{243}$
A: We can describe this using a four-state markov chain.  Define a matrix $A$ such that entry $a_{m,n}$ (the $m$'th row, $n$'th column entry) corresponds with the probability to move from state $n$ to state $m$.
We then have the transition matrix being:
$$A=\begin{bmatrix}0&\frac{1}{3}&\frac{1}{3}&\frac{1}{3}\\\frac{1}{3}&0&\frac{1}{3}&\frac{1}{3}\\\frac{1}{3}&\frac{1}{3}&0&\frac{1}{3}\\\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&0\end{bmatrix}$$
The probability you are interested in, $P(n)$ is calculated then as $P(n)=A^{n-1}\left[\begin{smallmatrix}1\\0\\0\\0\end{smallmatrix}\right]$
Using a calculator you can find specific values you are interested in, for example that $D$ is indeed $\frac{61}{243}$

Alternatively, due to the symmetry of the problem, we may instead simply describe this as a two-state markov chain, either using bat1 as the first state or using any different bat as the second state.  Like this, we have the transition matrix:
$$A=\begin{bmatrix}0&\frac{1}{3}\\1&\frac{2}{3}\end{bmatrix}$$
and we have
$$P(n)=A^{n-1}\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$$
As before, a calculator will again confirm the answer to $D$ is $\frac{61}{243}$.
A: For Your Option C,
Let the matches be P,Q,R,S
Now. The $B_1$ has been used in match P and is to be used in match S. Therefore It shouldn't be used in match R and obviously it can't bes used in match Q . For this, Probability is simply $\frac{2}{3}.\frac{1}{3}$ ($B_1  $ not used in match R and $B_1$ used in match S).
On the similar grounds, I'm getting $\frac{8}{81}$ for option D. Although I'm not pretty sure about it.
