You have to show that, for each $A$ fixed in the countable product $\sigma$-algebra, the map $x \to \mathbb{P}_x(A) = \mathbb{P}(S^x \in A)$ is measurable.
Since $S^x = S^0 + \underline{x}$, where $\underline{x} = (x, x, ..., x, x, ...)$, we have $\mathbb{P}_x(A) = \mathbb{P}_0(A - \underline{x})$. Setting $A_x = A - \underline{x}$, we have $\mathbb{P}_x(A) = \mathbb{P}_0(A_x) = \mathbb{E}_{\mathbb{P}_0}(\chi_{A_x})$.
Fact : If $(X, \mathcal{A})$ is a measurable space, $(\Omega,\mathcal{T}, \mathbb{P})$ is a probability space, and $f : X \times \Omega \to [0,+\infty]$ is a measurable map, then the map $x \to \mathbb{E}_{\mathbb{P}}(f(x, .))$ is measurable.
(if you want to prove it, it's an application of a Dynkin-class argument : prove it first for characteristic functions of measurable rectangles, and then extend the result in the usual way)
So, we are reduced to prove that the map $(x, \omega) \to \chi_{A_x}(\omega)$ is measurable. But this function is just the characteristic function of the set of all $(x, \omega)$ such that $\underline{x} + \omega \in A$, and since the map $(x, \omega) \to \underline{x} + \omega$ is measurable (exercise !), we are done.
Hint $1$: Hoeffding's bound is usually used on sums of independent random variables (usually Bernoulli or Rademacher). What could be the sums here? What are the independent random variables?
Hint $2$: Introduce some variable for the starting position and then try to figure out how $I_t$ can be expressed such that this starting position does not appear in the formula.
Hint $3$: Bound the probability that $|I_t|$ is large using Hoeffding.
Hint $4$: Recall the Union Bound.
Solution:
Let $X_i$ be the random variable which denotes the direction of the $i$-th step i.e. it is $-1$ with probability $1/2$ and $1$ with probability $1/2$. Let $s$ denote our starting point. As given in the exercise, we condition on the fact that $s + \sum_{i = 1}^n X_i = 0$. Hence $I_t$ is given by $I_t = s + \sum_{i = 1}^t X_i = \sum_{i = t + 1}^n X_i$ and has mean $0$. In particular, by the Hoeffding bound we have that
$$\text{P}(|I_t| \geq c \log n) \leq 2 \exp \left ( - \frac{(n - t)^2c^2\log^2n}{4(n - t)}\right) \leq 2 \exp \left ( - \frac{c^2\log^2n}{4}\right).$$
Using the Union-Bound we hence get
$$\text{P}(\max_t I_t - \min_t I_t \geq 2c \log n) \leq 2n \exp \left ( - \frac{c^2\log^2n}{4}\right) = \frac{2n}{n^{\frac{c^2\log n}{4}}}.$$
We conclude that we get $\max_t I_t - \min_t I_t = \mathcal{O}(\log n)$ with probability at least $1 - \frac{1}{\text{poly}(n)}$.
Best Answer
Let $k_{ij} = \mathbb E[T\mid X_0 = i, Y_0=j]$. So we want $k_{00}$. Noting the clear symmetry in both axes, we have the equations \begin{align*} k_{00}&=1+k_{10} \\ k_{10}&=1+\frac{1}{4}k_{00}+\frac{1}{4}k_{20}+\frac{1}{2}k_{11} \\ k_{20}&=1+\frac12k_{12}+\frac14k_{10} \\ k_{11}&=1+\frac12(k_{12}+k_{10}) \\ k_{12}&=1+\frac14(k_{11}+k_{20}+k_{22})\\ k_{22}&=1+\frac12k_{12} \end{align*} If you crunch through this system, you will get $\boxed{k_{00}=\tfrac{135}{13}}$.
For the second part, let $p_{ij}$ be the probability, conditional on $X_0=i$, $Y_0=j$, that $(X_T,Y_T)$ is one of $(\pm3,0),(0,\pm3)$. So we want $p_{00}/4$. Similar to above, we can get the system \begin{align*} p_{00}&=p_{10} \\ p_{10}&=\frac{1}{4}p_{00}+\frac{1}{4}p_{20}+\frac{1}{2}p_{11} \\ p_{20}&=\frac14+\frac12p_{12}+\frac14p_{10} \\ p_{11}&=\frac12(p_{12}+p_{10}) \\ p_{12}&=\frac14(p_{11}+p_{20}+p_{22})\\ p_{22}&=\frac12p_{12} \end{align*} Solving gives $p_{00}=\frac{4}{13}$, so $\boxed{\mathbb P(X_T=3,Y_T=0)=\tfrac{1}{13}}$.
In case you're interested, similar computations to the above gives $\mathbb P(X_T=3, Y_T=1)=\frac{3}{52}$ and $\mathbb P(X_T=3,Y_T=2)=\frac{3}{104}$.
Another aside: if you have a particular aversion to simultaneous equations, you might like to try some kind of martingale trick. This is especially appealing when you realise that the stopping time can be redefined as $T=\inf\left\{n\geq0\mid X_n^2+Y_n^2\geq9\right\}$. Why is this nice? Well, you can show that $M_n=X_n^2+Y_n^2-n$ is martingale, so by the optional stopping theorem, $$\mathbb E[M_T]=\mathbb E[M_0]=0\implies\mathbb E[T]=\mathbb E\left[X_T^2+Y_T^2\right].$$ We know that $X_T^2+Y_T^2\in\{9,10,13\}$. So we certainly have the bounds $9<\mathbb E[T]<13$. Unfortunately, I don't think we can glean much more information than this from martingale magic.