real analysis 

a) You need to check:
For every $\epsilon>0$ there is a $\delta>0$ so that if $|x-1|<\delta$ you get $|f(x)-f(1)|<\epsilon$.

So let $\epsilon>0$ and then take $\delta = \min(1,\frac19 \epsilon)$. If $|x-1|<\delta$ you have that
$$|x^2-1| < |(x-1)(x+1)|<|x+1|\delta=|x-1+2|\delta?(|x-1|+2)\delta < 2\delta + \delta^2$$
which is $?2\delta+\delta=3\delta$ since $\delta ?1$, whence $\delta^2 ?\delta$. Using this you get:
$$|f(x)-f(1)| = |3x^2+1 -(3\cdot1^2+1)| = 3|x^2-1| < 3\cdot (3\delta)=9\delta?9\frac19\epsilon =\epsilon$$
since $\delta?\frac19\epsilon$. This has then verified the $\epsilon$-$\delta$ condition.

b) Note that $f(x)=3x^2+1$ has a global minimum at $x=0$, where it take the value $1$. So if you take $A=(-1,1)$, which is open, you get $f(A) = [1,4)$, which is not open. So the function is not an open function.