Let $(S_0,T_0)=0$ and define $\{(S_n,T_n):n=0,1,2,\ldots\}$ by the transition probabilities
$$
\mathbb P((S_{n+1},T_{n+1}) = (i',j') \mid (S_n,T_n) = (i,j) = \begin{cases}
\frac14,& |i'-i| + |j'-j| = 1\\
0,& \text{otherwise}.
\end{cases}
$$
By symmetry,
$$
\mathbb P((S_1,T_1) = (1,0)) = \mathbb P((S_1,T_1) = (0,1)) = \mathbb P((S_1,T_1) = (-1,0)) = \mathbb P((S_1,T_1) = (0,-1)) = \frac14.
$$
For the distribution of $(S_2,T_2)$, there are three cases. First, the case where two steps are made in the same direction:
$$
\mathbb P((S_2,T_2) = (2,0)) = \mathbb P((S_2,T_2) = (0,2)) = \mathbb P((S_2,T_2) = (-2,0)) = \mathbb P((S_2,T_2) = (0,-2)).
$$
These probabilities are given by
\begin{align}
\mathbb P((S_2,T_2) = (2,0)) &= \mathbb P((S_2,T_2) = (2,0)\mid (S_1,T_1)=(1,0))\mathbb P((S_1,T_1)=(1,0))\\
&= \left(\frac14\right)^2\\
&= \frac1{16}.
\end{align}
Second, the case where one horizontal step is made and one vertical step is made:
$$
\mathbb P((S_2,T_2) = (1,1)) = P((S_2,T_2) = (-1,1)) = P((S_2,T_2) = (1,-1)) = P((S_2,T_2) = (-1,-1)).
$$
Since the steps could have been made in two different orders, these probabilities are $2\cdot\frac1{16}=\frac18$.
Third, the case where $(S_2,T_2)=(0,0)$. There are four ways this can happen, so the probability is $4\cdot\frac1{16}=\frac14$.
The distribution of $(S_3,T_3)$ may be found by a similar analysis - the probabilities will be multiples of $\frac1{4^3}=\frac1{64}$ depending on how many paths there are that end at a given point.
The event $E_{-2,-1} = \{S=-2,T=-1\}$ means that in the three steps you chose, one had to be down and two had to be to the left, in any order. So, the number of ways to choose the particular step when you went down was $\binom{3}{2}$.
At the same time, at each particular step, the probability of going down is $p_D = 1/4$, and the probability of going left is $p_L = 1/4$ each. Therefore,
$$
\mathbb{P}\left[E_{-2,-1}\right]
= \binom{3}{2} \cdot p_D \cdot p_L^2
= \binom{3}{2} \left(\frac{1}{4}\right)^3
$$
To find the marginal PMFs of $X$ and $Y$, you can find their joint PMF using a similar construction to the one above, and then sum across one of the variables, which will give you the PMF of the other one.
Best Answer
Hint: which is the expected value of a random variable uniformly distributed in the interval $[L-A,L+A]$?