Let $X_{t}=\sigma\cdot B_{t}$ ($\sigma\in\mathbb{R}$) be a Brownian bridge where $B_{t}$ is a standard Brownian bridge with fixed endpoints: $B_{T_{1}}=a$ and $B_{T_{2}}=b$ on the interval $[T_{1},T_{2}]$, and $m$ is a real constant.
How can I proove
$$\mathbb{P}\left(\exists t\in\left[T_{1},T_{2}\right]:\sup_{T_{1}\leq t\leq T_{2}}X_{t}>m\right)=\exp\left(-\frac{2\left(m-a\right)\left(m-b\right)}{\left(T_{2}-T_{1}\right)\sigma^{2}}\right)? $$ I can proove $\mathbb{P}\left(\exists t\in\left[0,1\right]:\sup_{0\leq t\leq1}B_{t}>a+(b-a)\cdot t;\;a,b\in\mathbb{R}\right)=\exp\left(-2ab\right)$, if $B_{t}$ is a standard Brownian bridge with $0$ startpoint and endpoint, but I couldn't deal with the asked problem.
To simplify things, I'll set $T_1=0$, $T_2=l$, and $\sigma=1$. Then the law of $\{X(t):0\le t\le l\}$ equals to that of $$ \left\{a+(b-a)\frac{t}{l}+\frac{l-t}{l}W\!\left(\frac{lt}{l-t}\right):0\le t\le l\right\}, $$ where $\{W(t)\}$ is a standard BM. Thus, for $m\ge a\vee b$, $$ \mathsf{P}(\exists t\in[0,l]:X(t)> m)=\mathsf{P}\!\left(\sup_{s\ge 0}\left(W(s)-\frac{m-b}{l}\cdot s\right)> (m-a)\right)=e^{-\frac{2(m-b)(m-a)}{l}}. $$ (See, e.g., Eq. 1.1.4(1) on page 256 here.)