2 dimensional symmetric random walk

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 $\boxed{\mathbb P(X_T=3,Y_T=0)=\tfrac{1}{13}}$.

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In case you're interested, similar computations to the above gives $\mathbb P(X_T=3, Y_T=1)=\frac{3}{52}$ and $\mathbb P(X_T=3,Y_T=2)=\frac{3}{104}$.

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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.

The answer above is entirely correct. Am just here with some simple code to give you confidence that $\mathbb{E}[T] = \frac{135}{13} $