Martingale Construction for Hitting Time on a Non-Symmetric Chain

A Markov chain on $\{0, 1, 2, 3\}$ has transition probabilities: $$p(1, 0) = \tfrac{1}{3}, \quad p(1, 2) = \tfrac{2}{3}, \quad p(2, 1) = \tfrac{1}{2}, \quad p(2, 3) = \tfrac{1}{2}.$$ State $0$ is absorbing and state $3$ is reflecting: $p(3, 2) = 1$.

Let $T = \inf\{n \ge 0 : X_n = 0\}$.

(a) Find a function $f : \{0,1,2,3\} \to \mathbb{R}$ and a constant $c > 0$ such that $M_n = f(X_{n \wedge T}) - (n \wedge T) \cdot c$ is a martingale. Use this to compute $E[T \mid X_0 = 2]$.

(b) Find a function $g : \{0,1,2,3\} \to \mathbb{R}$ such that $N_n = g(X_{n \wedge T}) - (n \wedge T) \cdot d$ is a martingale for an appropriate constant $d$, and use the optional stopping theorem to compute $E[T^2 \mid X_0 = 2]$. Hence find $\text{Var}(T \mid X_0 = 2)$.