Part (i): Use $$M_t := \frac{W_{t \wedge \tau(-b)} + b}{b}.$$
How to come up with this choice? Well, we are interested in $$\sup_{0 \leq t \leq \tau(-b)} W_t = \sup_{t \geq 0} W_{t \wedge \tau(-b)}.$$ We know that $(W_{t \wedge \tau(-b)})_{t \geq 0}$ is a martingale and that $W_{t \wedge \tau(-b)} \to -b$ as $t \to \infty$. This means that, in order to get a martingale $(M_t)_{t \geq 0}$ satisfying the assumption $M_t \to 0$ as $t \to \infty$, we have to shift the process. Adding $b$ yields a new martingale
$$\tilde{M}_t := W_{t \wedge \tau(-b)} +b \to 0 \qquad \text{as $t \to \infty$}.$$
By the definition of $\tau(-b)$, we have $\tilde{M}_t \geq 0$, i.e. $(\tilde{M}_t)_{t \geq 0}$ satisfies all the assumptions except $\tilde{M}_0=1$. Scaling the martingale with the factor $\frac{1}{b}$, we are done.
Part (ii): Consider the martingale $$M_t := \exp \left( \lambda W_t - \frac{\lambda^2}{t} \right) = \exp \left( \lambda \left[ W_t - \frac{\lambda}{2} t \right] \right).$$ (See also this question and this question.)
Brownian motion, Solution I
Since $W_t \sim N(0,t)$, we have
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) = \frac{2}{\sqrt{2\pi t}} \int_K^{\infty} x \exp \left( -\frac{x^2}{2t} \right) \, dx = \sqrt{\frac{2}{\pi}} \sqrt{t} \exp \left(-\frac{K^2}{2t} \right)$$
and therefore
$$\sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})=\infty.$$
Brownian motion, Solution II
Fix $K \geq 1$. Then, using that $W_t \stackrel{d}{=} \sqrt{t} W_1$,
$$\mathbb{E}(|W_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P}(|W_t|>K) = \mathbb{P}\left(|W_1| > \frac{K}{\sqrt{t}} \right) \stackrel{t \to \infty}{\to} 1.$$
Consequently,
$$\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|W_t| 1_{\{|W_t|>K\}})\geq 1.$$
(Non)Uniform integrability of $M_t := W_t^2-t$
For $K \geq 1$, we have by the triangle inequality
$$\mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) \geq \mathbb{P}(|M_t|>K) \geq \mathbb{P}(|W_t|^2 > K-t).$$
Using again the scaling property, we find
$$\mathbb{E}(|M_t| 1_{\{|W_t|>K\}}) \geq \mathbb{P} \left( |W_1|^2 > \frac{K-t}{t} \right). \tag{1}$$
Choosing $t = K/2$, we get
$$\begin{align*}\lim_{K \to \infty} \sup_{t \geq 0} \mathbb{E}(|M_t| 1_{\{|M_t|>K\}}) &\geq \liminf_{K \to \infty} \mathbb{E}(|M_{K/2}| 1_{\{|M_{K/2}|>K\}}) \\ &\stackrel{(1)}{\geq} \mathbb{P}(|W_1|^2 > 1) >0. \end{align*}$$
This shows that $(M_t)_{t \geq 0}$ is not uniformly integrable.
Best Answer
Following your train of thought, we can represent $M_t$ as a time-changed Brownian motion $\beta$ such that $\beta_{\langle M \rangle_t} = M_t$. Thus, $$ M_t - \frac{1}{2} \langle M \rangle_t = -\langle M \rangle_t \left( \frac{\beta_{\langle M \rangle_t}}{\langle M \rangle_t} + \frac{1}{2} \right) . $$ As $t \to \infty$, the second bracketed term tends to $1/2$ almost surely (since $\langle M \rangle_\infty = \infty$ a.s., we can treat this as our new time).