To find a sequence on $L^1$-norm equal to 2, converging a.e. to a function of $L^1$norm equal to 1.

For the first part, you may choose that $$g(x) = \left\{\begin{matrix}2\quad 0\leq x \leq \frac12\\0\quad \frac12 <x\leq 1\end{matrix}\right.$$ And for $n\geq2$ $$f_n(x) = \left\{\begin{matrix}2\quad 0\leq x \leq \frac12\\\quad0\quad \frac12<x \leq 1-\frac1n \\\quad n\quad 1-\frac1n <x\leq 1\end{matrix}\right.$$ It is easy to check that $(f_n)$ converges to g a.e. . However, $\Vert{f_n}\Vert = 2$ and $\Vert g \Vert=1$.
For problem(b), by applying Egroff's theorem, you can get such a subset E that $m(E)=1-\epsilon$,$m([0,1]\backslash E)=\epsilon$ and$(f_n)$ converges to $g$ uniformly on E.Then, you may write the integral as follows:$$\lim_n\int_0^1|f_n(x)-g(x)|=\lim_n(\int_E|f_n(x)-g(x)|+\int_{[0,1]\backslash E}|f_n(x)-g(x)|)$$
Claim:$$\lim_n\int_{[0,1]\backslash E}|f_n(x)-g(x)|=1$$
First observe that, $$\int_{[0,1]\backslash E}|g(x)| <\delta$$by the fact $m(E)=1-\epsilon$ and the absolute continuousness of L-intergral. And$$\int_{[0,1]\backslash E}|f_n(x)|\to 1$$ since $\int_{E}|f_n(x)-g(x)|\to 0 $ and $1\geq \int_{E}|g(x)| \geq 1-\delta$.
And the conclusion follows.