It’s always a good idea to look at some simple examples to try to get a feel for what’s going on. In (2), for instance, what happens if $f=g$?
For that matter, what happens in (1)? Since $f=g$ is a bijection from $[0,1]$ to $[0,1]$, there must be some $x\in[0,1]$ such that $f(x)=g(x)=1$, so that $(f+g)(x)=2\notin[0,1]$. If that’s not good enough $-$ after all, $f+g$ might still be a bijection between $[0,1]$ and some set $-$ consider two of the simplest bijections between $[0,1]$ and itself: $f(x)=x$, and $g(x)=1-x$. With that pair you can settle both (1) and (3) immediately, and with a little more thought, perhaps aided by looking at their graphs, you can settle (4) as well.
That leaves only (5), and you should already know that the composition of bijections is a bijection.