Let $Z=max(X,Y)$.

For the first subtask, keep a 2-D boolean array storing if the first player can win given a certain state. The states are number of stones remaining and number of stones player is allowed to take on this turn. This array can be computed using DP.

Time Complexity: $\mathcal{O}(Z^3)$

For the second subtask, say that there are $N$ stones currently. The first player can win if they are allowed to take $A$ stones on this turn. Then clearly, the player can win if they are allowed to take $B$ stones on this turn and $B\ge A$. So store the least amount of stones the first player must take to guarantee a win if there are currently $N$ stones and call this $dp[N]$. A DP recurrence can be found involving this array: if $M$ is the largest number (less than $N$) such that $dp[M]>N-M+K$, then $dp[N]=N-M$. (Also, $dp[0]=\mathrm{\infty }$.)

Time Complexity: $\mathcal{O}(Z^2)$

For the third subtask, finding each $dp[N]$ in the second subtask can be sped up. Previously, it took $\mathcal{O}(N)$ to compute $dp[N]$. Using data structures such as an RMQ segment tree, this may be sped up to $\mathcal{O}(\log N)$.

Time Complexity: $\mathcal{O}(Z\log Z)$

For the fourth subtask, first note the following fact: For any $M$ with $dp[M]\ne M$, let $N$ be the largest number less than $M$ such that $dp[N]=N$. Then $dp[M]=dp[M-N]$. This claim can be proved by strong induction on $M-N$.

Consider an arbitrary $N_1$ with $dp[N_1]=N_1$ and $N_1\ge K+2$ (it is not hard to see that $dp[N]=N$ for all $1\le N\le K+2$). Let $N_3$ be the smallest number larger than or equal to $N_1-K$ with $dp[N_3]=N_3$ and $N_2$ be defined as $N_1+N_3$. Then $N_2$ is the least number larger than $N_1$ such that $dp[N_2]=N_2$.

Proof: It can be seen from the earlier claim and from the fact that $dp[N_2-N_1]=N_2-N_1$, that for any $N$ with $N_1<N<N_2$, that $dp[N]\ne N$ and also $dp[N]=dp[N-N_1]\le N_3-(N-N_1)+K=N_2-N+K$. Now for any $N$ with $0<N\le N_1$, $dp[N]\le N=N-(N_1+N_1-K)+K\le N-(N_1+N_3)+K=N-N_2+K$. So the largest number $M$ less than $N_2$ such that $dp[M]>N_2-M+K$ is $0$. So $dp[N_2]=N_2$.

So using this, all $N$ such that $dp[N]=N$ can be computed. There are approximately ${\log }_{2}Z$ such $N$s between $K+2$ and $Z$. For each $N$ with $dp[N]=N$, compute the number of $M$s such that $dp[M]=1$ using the first claim. Then, using these values, the final answer can be computed using the first claim.

Time Complexity: $\mathcal{O}(\log Z)$