1. what are symmetric expressions...
As noted in the first paragraph, symmetric expressions are those expressions in α and β which do not change by interchanging α and β.
An expression in $\,a,b\,$ can be written as $\,f(a,b)\,$. Interchanging $\,a\,$ and $\,b\,$ gives the expression $\,f(b,a)\,$. By definition, the original expression is symmetric in a and b if and only if $\,f(a,b)=f(b,a)\,$ for all $\,a,b\,$.
$\,f(a,b)=a^2+b^2+ab+1\,$ is symmetric in $\,a,b\,$ because $\,f(b,a)=b^2+a^2+ba+1$ $=a^2+b^2+ab+1=f(a,b)\,$ for all $\,a,b\,$.
$\,f(a,b)=a^2-2b^2\,$ is not symmetric in $\,a,b\,$ because $\,f(b,a)=b^2-2a^2 \ne a^2-2b^2=f(a,b)\,$. Even though it is possible that $\,f(a,b)=f(b,a)\,$ for some $\,a,b\,$ e.g. $\,f(1,-1)=f(-1,1)\,$, the equality does not hold true for all values e.g. $\,f(1,0)=1 \ne -2 = f(0,1)\,$, so the expression is not symmetric.
...in Quadratic Equations.
It's not about "symmetric expressions in quadratic equations", but rather "symmetric expressions in the roots of quadratic equations".
The first paragraph refers to the roots α, β of an equation. Since there are two roots, and assuming the context is algebraic, the equation must be a quadratic, such as $\,(x-\alpha)(x-\beta)=0\,$. Expanding and collecting, the quadratic can also be written as $\,x^2-px+q=0\,$, where $\,p=\alpha+\beta\,$ and $\,q=\alpha\beta\,$.
2. Why they are generally expressed in terms of $\alpha + \beta$ and $\alpha \beta$?
Because it is possible: by the fundamental theorem of symmetric polynomials, any polynomial expression in $\,\alpha,\beta\,$ which is symmetric in $\,\alpha,\beta\,$ can be expressed as a polynomial in the elementary symmetric polynomials, which in the case of two variables are just $\,\alpha+\beta\,$ and $\,\alpha \beta\,$.
Because it is convenient: by Vieta's formulas, the elementary symmetric polynomials in the roots of a polynomial are directly related to the coefficients of the equation. In the simple case of a quadratic $\,x^2-px+q=0\,$, Vieta's formulas are precisely $\,\alpha+\beta=p\,$ and $\,\alpha\beta=q\,$ from the previous step.
3. The given examples of the symmetric expressions (as given above 1, 2 and 3) aren't quadratic equations really, they seem to like in 2 variable. What are they?
Those are symmetric expressions in the roots of a quadratic equation.
Taking the first one for example, the complete statement of the problem would be something like:
$$
\style{font-family:inherit}{\text{Find the value of the expression}} \;\alpha^2+\beta^2\; \style{font-family:inherit}{\text{where}} \;\alpha,\beta\; \style{font-family:inherit}{\text{are the roots of}} \;x^2 - px+q=0\,.
$$
Using the hint, and using that $\,\alpha+\beta=p\,$ and $\,\alpha \beta=q\,$, the expression can be calculated as:
$$
\alpha^2+\beta^2\color{red}{+2\alpha\beta-2\alpha\beta} = (\alpha+\beta)^2 - 2 \alpha\beta = p^2 - 2q
$$
It would be technically possible, of course, to actually solve the quadratic, find the roots $\,\alpha\,$ and $\,\beta\,$ explicitly, then calculate $\,\alpha^2+\beta^2\,$ the long and hard way. However, since the expression was symmetric, it was possible to calculate it a lot more quickly without having to actually find the individual values of $\,\alpha,\beta\,$.
You're quite right that there cannot be a cubic term in a quadratic form -- for example, as written $Q(0,0,2)\ne 4Q(0,0,1)$, contradicting the definition.
It looks to me like the apparently cubic term is just a typo in the problem. It should be $x_3^2$, not $x_3^3$.
If you are to hand in written homework, I would simply start it with:
I assume the form was supposed to be $8x_1^2+2x_2^2+3x_3^2+8x_2x_3$ ...
Best Answer
The first one is a simple quadratic in standard form. The rest of them are just variations on the vertex form of a quadratic. Specifically, they are of the form $\alpha (x-\beta)^2 + \gamma,\alpha, \beta,\gamma \in\mathbb{R}$.
To express the given equation in the form $p-q(x+r)^2,$ notice that $q=2$ (just expand the square and use the same coefficient-matching method you demonstrated). By the same method, you should get $r=-1$ and $p=15$. So the equation is $15-2(x-1)^2.$