$(X_t)_{t \ge 0}$ is a process adapted to $(\mathcal F_t)_{t \ge 0}$ with continuous trajectories and such that $X_0=1, X_t$ is non-negative, $X_t ^4$ is a martingale with respect to $(\mathcal F_t)$ and $\limsup _{t \to \infty} X_t = +\infty$. Let for $a>0$: $$\tau_a=\inf \{ t>0: X_t =a\}.$$ Show that for $0<b<1<a$ $$\mathbb P(\tau_a < \tau_b)=\frac{1-b^4}{a^4-b^4}.$$
I have absolutely no idea for this task. $$\mathbb P(\tau_a < \tau_b)=\mathbb P(\inf \{ t>0: X_t =a\}-\inf \{ t>0: X_t =b\}<0)$$ I think you need to make clever use of the stopping time properties and the facts about $X_t$ given in the problem, but I didn't get anything sensible.
Let $\rho=\min(\tau_a, \tau_b)$. Observe that $\rho$ is a stopping time, since $$ \rho=\inf \{ t>0: X_t \in \{ a,b \} \}. $$
By the optional stopping theorem, $$ \mathbb E(X_{\rho}^4)=\mathbb E(X_0^4)=1. $$ On the event $\tau_a<\tau_b$ we have that $X_{\rho}=X_{\tau_a}=a$, and likewise on the event $\tau_a>\tau_b$ we have that $X_{\rho}=b$. The event $\{\tau_a=\tau_b\}$ occurs with probability $0$. Thus $$ \mathbb E(X_{\rho}^4)=a^4\mathbb P(\tau_a<\tau_b)+b^4\mathbb P(\tau_a>\tau_b),\qquad 1=\mathbb P(\tau_a<\tau_b)+\mathbb P(\tau_a>\tau_b). $$ Substituting $\mathbb E(X_{\rho}^4)=1$ therefore yields the matrix equation $$ \begin{pmatrix}a^4 & b^4 \\ 1 & 1 \end{pmatrix}\begin{pmatrix}\mathbb P(\tau_a<\tau_b)\\ \mathbb P(\tau_a>\tau_b)\end{pmatrix}=\begin{pmatrix}1 \\ 1\end{pmatrix} $$ whereupon $$ \begin{pmatrix}\mathbb P(\tau_a<\tau_b)\\ \mathbb P(\tau_a>\tau_b)\end{pmatrix}=\begin{pmatrix}a^4 & b^4 \\ 1 & 1 \end{pmatrix}^{-1}\begin{pmatrix}1 \\ 1\end{pmatrix}=\frac{1}{a^4-b^4}\begin{pmatrix}1 & -b^4 \\ -1 & a^4 \end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}. $$ Thus $$ \mathbb P(\tau_a<\tau_b)=\frac{1-b^4}{a^4-b^4},\qquad \mathbb P(\tau_a>\tau_b)=\frac{a^4-1}{a^4-b^4}. $$