All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
From GC, $\alpha = 1.885$ to 3 d.p.

$x^{6}-3x^{4}-7=-7$

$x^{4}(x^{2}-3)=0$

$x=0$ or $x^{2}=3$

$x=0$ or $x=\pm{\sqrt{3}}$

Since $\beta > 0$, $\beta =\sqrt{3}$

(ii)
Using the GC, $\mathrm{required~ answer} = -0.597$ to 3 d.p.

(iii)
Area
$= - \int_{0}^{\sqrt{3}} f(x)+7 dx$

$= - \int_{0}^{\sqrt{3}} x^{6}-3x^{4} dx$

$= -(\frac{x^{7}}{7}-\frac{3x^{5}}{5})\biggl|_0^{\sqrt{3}}$

$= \frac{54}{35} \sqrt{3}$

(iv)
$f(-x)=(-x)^{6}-3(-x)^{4}-7=x^{6}-3x^{4}-7=f(x)$
Therefore, $f(-x)=f(x)$ and f is an even function.

Since the coefficients of $f(x)$ are all real, by conjugate root theorem, all complex roots occur in conjugate pairs. And since it is an even function, if $a+bi$ is a root, then $-(a+bi)$ is also a root.
Thus, we have 2 real roots, $\pm \alpha$ and 4 complex roots, $\pm a \pm bi$, where $a, b \in \mathbb{R}$