Let $X_1(t)=e^{B(t)}$ and $X_2(t)=e^{-B(t)}$ where $B(t)$ is the standard brownian motion and $\{G_t\}$ is the filtration generated by the brownian motion. Determine what kind of martingale $X_1(t)$ and $X_2(t)$ are.
I'm fairly certain this comes out trivial as they're both $\{G_t\}$-measurable, and convex functions, so by jensens inequality for martingales they're both submartingales.
Is there a way that could result in a stronger statement for one of these?
There are at least four ways to answer this question.
As you pointed out in your answer, use the fact that if $M_t$ is a martingale and $f(\cdot)$ is convex then $f(M_t)$ is a submartingale.
Apply Jensen's inequality directly to $\mathbb{E}\left[e^{B_t}\right]$. By "conditional version" of Jensen's inequality we have $$ \mathbb{E}\left[e^{B_t}|\mathcal{F_s}\right] > e^{\mathbb{E}[ B_t|\mathcal{F_s}]}=e^{B_s} $$ since $f(\cdot)$ is strictly convex.
As Surb suggested in his comment, use Ito's Lemma to obtain $$ e^{B_t}= 1 + \int_0^t e^{B_s}dB_s + \frac{1}{2}\int_0^t e^{B_s}ds,$$
(My personal favorite) Use the moment generating function of $B_t$. We know that $B_t\sim \mathcal{N}(0,t)$ and so $$\mathbb{E}\left[e^{uB_t}\right]=e^{\frac{1}{2}u^2t}$$ From this we conclude that, for any $u\in\mathbb{R}$, we have $$\mathbb{E}\left[e^{uB_t}|\mathcal{F_s}\right]=e^{uB_s}\mathbb{E}\left[e^{u(B_t-B_s)}|\mathcal{F_s}\right]=e^{uB_s}\mathbb{E}\left[e^{u(B_t-B_s)}\right]=e^{uB_s}e^{\frac{1}{2}u^2(t-s)}>e^{uB_s} $$