If the $\displaystyle L_1$-limit of the sequence $\displaystyle f_n(t) = \frac1{\sqrt n}e^{-\frac1n(t-n)^2}$ exists, then it must be equal to the pointwise limit, which is the zero function. So all you need to do is to check whether $\displaystyle \|f_n\|_1 = \displaystyle\int_0^\infty \!\! f_n(t)\,dt \to0$ as $\displaystyle n\to\infty$. In fact, if you make the substitution $\displaystyle u = \frac1{\sqrt n}(t-n)$, you see that $\displaystyle \|f_n\|_1 = \displaystyle\int_{-\sqrt n}^\infty e^{-u^2}du \to\sqrt\pi$, and so the sequence does not converge in the $\displaystyle L_1$-norm.
I agree, the pointwise limit is $\displaystyle e^{2t} $. To show that this is the limit in the sup norm, you need to show that $\displaystyle e^n x_{n} \to e^{2t}$ uniformly on [0,1]. So look at the difference $\displaystyle e^n x_{n} - e^{2t}$ and see if that can be made small for all $\displaystyle t\in[0,1]$, provided that $\displaystyle n$ is large enough.