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.
Consider the polynomial $z-i$ . . .
It's good to remember the reason the complex conjugate theorem is true in the first place: the map $z\mapsto \overline{z}$ is an automorphism of the field $\mathbb{C}$, and fixes the subfield $\mathbb{R}$ (this is a fancy way of saying $\overline{r}=r$ if $r$ is real). Thus - defining the "conjugate" $\overline{p}$ of a polynomial (with coefficients in $\mathbb{C}$) to be the polynomial whose coefficients are the conjugates of the corresponding coefficients of $p$ - we have the following:
If $p$ is any polynomial and $z$ is a root of $p$, then $\overline{z}$ is a root of $\overline{p}$.
If the coefficients of $p$ are from $\mathbb{R}$, then $\overline{p}=p$.
This second fact is no longer true for polynomials with coefficients from outside $\mathbb{R}$!
Note that this generalizes in a natural way to:
Suppose $F$ is any subfield of $\mathbb{C}$, $\varphi$ is a field automorphism of $\mathbb{C}$ which fixes $F$, and $p$ is a polynomial with coefficients from $F$. Then if $z$ is a root of $p$, so is $\varphi(z)$.
This is the (fine, "a") first step towards Galois theory . . .
Best Answer
Let's go with yes. It's a bit complicated to straighten several things out here though.
What's the problem?
Firstly, we have to address the question of what $x^i$ means. There are a couple choices.
The easiest choice is of course to restrict to $x$ positive and real, in which case we can define $x^i$ as $e^{i\log x}.$
The other choice is to replace $x$ with $e^z$ with $z$ complex and solve for roots of the equation $$\sum_{n} a_n e^{izn},$$ and say that the equation has roots at $0$ based on whether the polynomial has a constant term or not.
Why is this a problem?
Well how do we define $(-1)^i$? Well $(-1)=e^{\pi i}=e^{3\pi i}$, so $(-1)^i$ should be $e^{-\pi}=e^{-3\pi}$, except wait, those two aren't equal. Thus we run into a problem with complex exponents. See the complex logarithm for more details. (Also there are other ways to resolve this problem, but I chose what I think are the two simplest above).
Solving your equation
Your "polynomial" isn't very interesting to solve, since if the polynomial is $$\sum_{n} a_n x^{in},$$ we can just solve the normal polynomial $$\sum_n a_n y^n$$ and then solve the equations $x^i = y$.
A much more interesting problem (probably, I don't know)
Now you mentioned that you didn't think it made sense to define polynomials with complex exponents, since you couldn't order the exponents. While it's true that there's no natural order for the exponents, there's no reason that they don't make sense. Perhaps it's best to restrict to Gaussian integers though to at least preserve some semblance of sanity.
We could define the "degree" of such a polynomial to be the largest norm among the exponents in the monomials, and ask if it's possible to find the solution to equations with "degree" at most 2.
For example, can we solve $$x^2 + x^{1+i} + x^{2i}?$$
I don't know, I haven't thought about it much, I just wanted to point out that there's no reason you can't ask the question, and that it might be more interesting than the question you did ask.
One approach to start trying to solve such an equation is to let $y=x^i$ again, so we get a two variable equation $$x^2+xy+y^2,$$ defining a curve, and then find the points on this curve that satisfy the constraint that $x^i=y$ for some appropriate solution to the problem of raising numbers to complex powers.