For 1: It's not only common, it's always the case: if you have a polynomial with real coefficients, then the "complex roots" come in what are called "conjugate pairs: if $a+bi$ is a root, with $a,b$ real numbers and $b\neq 0$, then $a-bi$ is also a root.
The explanation has to do with something called "complex conjugation". Complex conjugation is an operation on complex numbers that sends the complex number $a+bi$ to the complex number $a-bi$. This is denoted with a line over the number, that is, $\overline{z}$ denotes the complex conjugate of $z$:
$$\overline{a+bi} = a-bi,\qquad a\text{ and }b\text{ real numbers.}$$
Conjugation respects sums and products:
$$\overline{z+w} = \overline{z}+\overline{w}\text{ and }\overline{z\times w}=\overline{z}\times\overline{w}\text{ for all complex numbers }z\text{ and }w.$$
Also, a number $a+bi$ is real (that is, has $b=0$) if and only if $\overline{a+bi} = a+bi$.
Now suppose you have a polynomial with real coefficients,
$$p(x) = \alpha_nx^n + \cdots + \alpha_1x + \alpha_0$$
and that there are real numbers $a$ and $b$ such that $a+bi$ is a root of this polynomial. Plugging it in, we get $0$:
$$p(a+bi) = \alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0 = 0.$$
Taking complex conjugates on both sides, and applying the properties mentioned above (complex conjugate of the sum is the sum of the complex conjugates; complex conjugate of the product is the product of the complex conjugate; complex conjugate of a real number like $\alpha_i$ is itself) we have:
$$\begin{align*}
0 &= \overline{0}\\
&= \overline{\alpha_n(a+bi)^n + \cdots + \alpha_1(a+bi) + \alpha_0}\\
&= \overline{\alpha_n}\overline{(a+bi)}^n + \cdots + \overline{\alpha_1}\overline{(a+bi)} + \overline{\alpha_0}\\
&= \alpha_n(a-bi)^n + \cdots + \alpha_1(a-bi) + \alpha_0\\
&= p(a-bi).
\end{align*}$$
So if $p(a+bi)=0$, then $p(a-bi)=0$ as well.
Complex roots don't correspond in any reasonable way to "maxima and minima" of the polynomials. They correspond to irreducible quadratic factors. It's hard to visualize, from a graph of the polynomial in the real axis, where (or even whether) it has complex roots.
The polynomial you tried, $3x^6+4x^4$, can be written as
$$3x^6 + 4x^4 = x^4(3x^2 + 4).$$
The product is equal to $0$ if and only if one of the factors is $0$, so either $x=0$ or $3x^2+4 = 0$; for the latter, you would need $x^2 = -\frac{4}{3}$, which is impossible with real numbers; that is, $3x^2 + 4$ is an irreducible quadratic, which is where the complex roots (in this case, purely imaginary) come from. Since you want a (complex) number whose square is $-\frac{4}{3}$, one possibility is to take a real number whose square is $\frac{4}{3}$, namely $\frac{2}{\sqrt{3}}$, and then multiply it by a complex number whose square is $-1$. There are two such numbers, $i$ and $-i$, so the two complex roots are $\frac{2}{\sqrt{3}}i$ and $\frac{2}{\sqrt{3}}(-i) = -\frac{2}{\sqrt{3}}i$.
Each distinct irreducible quadratic factor will give you a pair of conjugate complex roots.
As a consequence of the Fundamental Theorem of Algebra, every polynomial with real coefficients can be written as a product of polynomials that are either degree $1$, or irreducible quadratics.
Might as well mention one application here: one reason people were interested in knowing that the Fundamental Theorem of Algebra was true (that every real polynomial could be written as a product of linear and irreducible quadratic polynomials) was to ensure that the method of partial fractions would always be available to solve an integral of a rational function (a function given as the quotient of a polynomial by another polynomial).
Other uses are too many to mention, but as Yuval mentions, this question and its answers may get you started.
Best Answer
No, they are not.
For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical.
And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the same roots, yet they are not identical.
Again, depending on what you mean by "the same roots", $x^3+x$ and $x^3+2x$ both only have one real root, yet they are not the same.
However, if two monic polynomials have the same roots, with the same multiplicities, over some algebraicaly closed field (like the complex numbers $\Bbb C$) then yes, they are identical.