What follows is a standard solution using the argument principle. There are many clever approaches that will work, by choosing the right places to use Rouché's theorem and using some inequalities, but this general approach almost never fails.
Argument principle: The argument principle states that over a closed curve, the number of zeros minus the number of poles is the change in argument over the curve divided by $2\pi$. This requires of course that there are no zeros or poles on the boundary. We can write it as $$2\pi(n-m)=\Delta_C \arg(f).$$
To use this, first note that there are no real roots with positive real part since all the coefficients are positive. Since all the coefficients are real, every root in the first quadrant will correspond to a root in the fourth quadrant, so we need only consider the roots in the first quadrant.
Now, notice that there are no roots on the imaginary axis since
$$(ix)^4+8(ix)^3+3(ix)^2+8(ix)+3=x^4-8ix^3-3x^2 +8ix+3=0$$ implies that both $$x^4-3x^2+3=0,\text{ and } 8x-8x^3=0$$ which is impossible.
Consider the contour which is a pizza slice of the right quarter circle or Radius $R$ in the first quadrant. (That is, start at zero, go along the positive real axis until $R$, and then follow the circle or radius $R$ until the imaginary axis, and then return back to the origin) We need only find the change in argument of the polynomial along this curve to know how many zeros are inside. (Since it has no poles.)
Computing the change in argument: We break this down into the three segments.
Real line: On the real line, our polynomial it is always positive, so the change in argument is zero.
Circular arc: On the circular arc, when we take $R$ larger and larger, the first term $z^4$ will dominate so that the change in argument there is the same as it is for $z^4$ in the limit. This gives a change in argument of $2\pi$.
Imaginary axis The imaginary axis is hardest to deal with. However, since the real part is $$x^4-3x^2+3=\left(z^2-\frac{3}{2}\right)^2+\frac{3}{4}$$ which is always positive, we see that it cannot leave the right half plane, and cannot wind around the origin. Hence only the final point and initial point matter, so we see the change in argument is $0$.
Hence as $R\rightarrow \infty$, we see that the change in argument is $2\pi$ so that there is exactly 1 zero. This means that there are two zeros in the right half plane.
Hope that helps,
To the last question, traditionally if one does not qualify the coefficients directly or from context, rational or integer coefficients are assumed, so the polynomial is a real function.
The calculations for the case with complex roots should be the same as with the real roots, just take $b,c$ complex instead of real, with $c=\bar b$.
$f(x)=(x-a)(x-b)(x-c)$, $f'(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$, the tangent at $x_0=\frac12(a+b)$ is
$$
t(x)=f(x_0)+(x-x_0)f'(x_0)=(x_0-a)(x_0-b)(x_0-c)+(x-x_0)(x_0-a)(x_0-b)
=(x-c)(x_0-a)(x_0-b)
$$
so it has a root at $x=c$.
However, you would get the third root more quickly by using that $-(a+b+c)$ is the quadratic coefficient of the polynomial.
Best Answer
If $a=b$, then write $k_1 e^{ax}-k_2e^{bx}$ as $ke^{ax}$.
If $k$ and $a$ are opposite sign, or if either is 0, then $x=ke^{ax}$ has exactly one solution.
Otherwise, $$x=ke^{ax}$$ if and only if $$ax=kae^{ax}.$$
Let $y=ax$. Then this is true of and only if $$y=kae^y,$$ that is, $e^y=my$ where $m=\frac{1}{ka}$. It's easy to characterise when this has 0, 1 or 2 solutions, and that gives you the values of $k$ and $a$ which give rise to 0, 1 or 2 solutions.
If $a\neq b$, then the cases $k_1=0$ or $k_2=0$ have been covered above.
Assume for now that $k_1,k_2>0$, and let $x=y+q$. Then $$x=k_1e^{ax}-k_2e^{bx}$$ if and only if $$y+q=k_1e^{ay}e^{aq}+k_2e^{by}e^{bq}\\=k_1e^{aq}\left(e^{ay}+\frac{k_2}{k_1}e^{(b-a)q}e^{by}\right).$$ Letting $$q=\frac{\ln(k_1/k_2)}{b-a},$$ this becomes $$\frac{y+q}{k_1e^{aq}}=e^{ay}-e^{by}.$$
Then, letting $z=\frac{y}{b-a}$, this becomes $$mz+c=e^{dz}\left(e^z-1\right),$$ where $m$, $c$ and $d$ are expressions in $a$, $b$, $q$, $k_1$ and so on.
Now, it's a matter of characterising the tangent lines to $e^{dz}\left(e^z-1\right)$, and translating those characterisations into regions in $(k_1,k_2,a,b)$ space. This almost-one-parameter problem is much more manageable than the original four-parameter problem.
If $k_1<0$ and $k_2<0$ you can proceed similarly to the above. If $k_1$ and $k_2$ have opposite sign, you can again proceed as above, but you'll need to characterise tangents to $e^{dz}\left(e^z+1\right)$