So, let ε>0 be given. Let |x-1|<δ. Now note that |f(x)-3| = 3|x-1| < 3δ.
Now we want this to be less than ε, so if we pick δ = ε/3, then |f(x)-3|<ε.
Thus, the limit of f(x) as x->1 is 3.
QED.
I hope you understand the underlying principle of the definition: if we choose x closer to 1, then f(x) will be closer to 3. The idea of the epsilon/delta proof is to let epsilon be any number and to then find a delta such that |x-1|<δ implies |f(x)-3|<ε. By letting epsilon be any number (greater than zero), the proof that follows works for all numbers (greater than zero).
Note also the importance of the implication. We assume |x-1|<δ and then show that from that assumption it follows that |f(x)-3|<ε.
Now, adressing your questions:
The value of delta is indeed an expression involving epsilon. It is therefore sometimes denoted by δ_ε or even δ(ε). This proof is indeed sufficient to show that the limit of f(x) is 3. This is because we defined the limit to be 3 if the statement
∀ε >0 ∃δ>0 s.t |x-1|<δ => |f(x)-3|<ε
holds. We've shown it to be true, therefore the limit is 3.