Finding parameters for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ that satisfy a condition. 1.find a for $f(x) = x^2 + 2(m − a)x + 3am −2 = 0$ such that for every m real, f has real roots
2.find m such that for every a real, f has real roots
My ideea is to demonstrate that $\delta=4(m-a)^2 - 4(3am-2) = 4(m^2-5am+a^2+2) >0$
But I have no ideea how to do this.
 A: As you correctly remark, it is enough that
$$\delta=4(m^2-2am+a^2)-4(3am-2)\ge0\iff  a^2-5ma+m^2+2\ge0$$
Now look at the discriminant of the new quadratic in $\;a\;$ you got:
$$\delta_a:=25m^2-4m^2-8=21m^2-8$$
We'd like to check when $\;\delta_a\le0\;$ so that that $\;\delta\ge0\;$. This will be true for all $a$ iff 
$$m^2<\frac8{21}\iff |m|\le\frac{2\sqrt2}{\sqrt{21}}$$
Taking into account that the quadratic $\;\delta\;$ is symmetric with respect to both $\;a,m\;$ we'll get the same result if we fix $\;a\;$ and calculate $\;\delta_m\;$ instead: $f(x)$ will have real roots for all $m$ iff $$|a|\le\frac{2\sqrt2}{\sqrt{21}}$$
A: A parabola $Ax^2+Bx+C$ with $A>0$ attains its minimum at $x_0 = -\frac{B}{2A}$, with value $y_0 = C-\frac{B^2}{4A}$. First the first question, this means that the minimum $\mu_a$ of $(m^2−5am+a^2+2)$ is always positive, whatever $m$. You can expect some solutions: if $a=0$, this quantity is equal to $\mu_0=m^2+2$, positive for every $m$.
The minimum value in general  is $\mu_a =a^2+2 -\frac{(5a)^2}{4}=\frac{8-21a^2}{4}$. If $|a|\le \sqrt{\frac{8}{21}}$, then the quantity $\delta(m,a)$ is positive. 
A verification: if $a=0$, the discriminant of $f_0(x) = x^2+2mx−2$ is $\delta_0 = 4m^2+8$ which is always positive, for all $m$.
You can use the same approach the other way round for the second question. But you really don't need:   $\delta(m,a) =  \delta(a,m)$, the function is symmetric with respect to the two variables. So you should find a quite similar result for $m$.
