Stochastic Ito Formula

Question

Let B be standard Brownian Motion, started at $0$, $X = (X_{t})_{t\geq0}$ a non-negative stochastic process solving:

\begin{equation} dX_{t} = \frac{1}{X_{t}}dt + dB_{t}, (X_{0} = 0)\end{equation}

For $F(t,x) = e^{-t}x^{2}$ for $t\geq0$ and $x\in\mathbb{R}_{+}$.

(1) Apply It$\hat{o}$'s formula to $F(t, X_{t})$ for $t\geq0$ and determine a continuous local martingale $(M_{t})$ starting at $0$ and a continuous variation process $(A_{t})$ such that $F(t, X_{t}) = M_{t}+A_{t}$

(2) Show $M_{t}$ in (1) is a martingale and compute $[M, M]_{t}$

(3) Compute $\mathbb{E}\tau$ for stopping time: $\tau = inf\{t\geq0: X_{t} = 1\}$

My attempt:

(1) \begin{align}F(t, X_{t}) &= F(0, X_{0}) - \int_{0}^{t}e^{-s}X_{s}^{2}ds + 2\int_{0}^{t}e^{-s}X_{s}dX_{s} + \frac{1}{2}\int_{0}^{t}2e^{-s}d_{s}\\ &= - \int_{0}^{t}e^{-s}X_{s}^{2}ds+ 2\int_{0}^{t}e^{-s}ds + 2\int_{0}^{t}e^{-s}X_{s}dB_{s} + \int_{0}^{t}e^{-s}ds \end{align}

(2) $M_{t} = 2\int_{0}^{t}e^{-s}X_{s}dB_{s}$ for $t\geq0$

$_{t} = 4\int_{0}^{t}e^{-2s}X_{s}^{2}ds$

I am stuck at this point.