Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$ 
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.


My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$.  This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.

I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?

Your attempt is mostly correct, apart from a few fairly minor issues regarding an open set being used, and there's some work you didn't need to do. First, as you pointed out, the inequality can be rewritten as
$$f(x) = \left(2x-\frac{a}{2}\right)^2 + \frac{3a^2}{4} - 5a + 4 \gt 0 \tag{1}\label{eq1A}$$
For any fixed value of $a$, then $\frac{3a^2}{4} - 5a + 4 = c$ is a constant. Also, with $g(x) = 2x-\frac{a}{2}$, then \eqref{eq1A} can be rewritten as
$$f(x) = g^2(x) + c \gt 0 \tag{2}\label{eq2A}$$
To have \eqref{eq2A} being true $\forall \, x \in (0,2)$, with
$$m = \min_{x\in[0,2]}g^2(x) \tag{3}\label{eq3A}$$
requires that $m + c \ge 0$ where $g(0) = 0 \lor g(2) = 0 \lor m \neq 0$ (i.e., points $1$ and $2$ below), and $m + c \gt 0$ otherwise (i.e., where $g(x) = 0$ for an $x \in (0,2)$, as described in point $3$ below). The first case is because the minimum value $m$ is not actually reached on the open set $(0,2)$, i.e., $g^2(x) \gt m \; \forall \; x \in (0,2)$, so $g^2(x) + c \gt 0$ means that $m + c \ge 0$.
To determine this minimum value $m$ in \eqref{eq3A} involves $3$ basic cases:

*

*$g(x) \lt 0 \; \forall \; x \in (0,2)$, which is your case $1$, with $m$ coming from the upper limit value of $g(x)$, i.e., $m = (4 - \frac{a}{2})^2$.

*$g(x) \gt 0 \; \forall \; x \in (0,2)$, which is your case $2$, with $m$ coming from the lower limit value of $g(x)$, i.e., $m = (-\frac{a}{2})^2$.

*$g(x)$ having both negative and positive values (so $0$ as well), which is basically your case $3$ (except you should've used $\gt$ instead of $\ge$ to avoid overlapping with your previous cases, as well to ensure that $g(x) = 0$ for some $x \in (0,2)$), with $m = 0$.

Regarding your analysis of these cases, note that $g(x)$ is a strictly increasing function, with its upper limit value of $4 - \frac{a}{2}$ being $4$ more than its lower limit value of $-\frac{a}{2}$ (and, correspondingly, the lower limit is $4$ less than the upper limit). Thus, for example, with your case $1$, having $4 - \frac{a}{2} \le 0 \; \to \; 4 \le \frac{a}{2} \; \to \; a \ge 8$, i.e., a stronger condition than your $a \ge 0$, so this didn't need to be mentioned. Further, as I explained in my point $1$, you only needed to deal with handling $4 - \frac{a}{2}$.
Similarly, with your case $2$, having $a \le 0 \; \to \; -\frac{a}{2} \ge 0$ means that $4-\frac{a}{2} \ge 4$, which is a stronger condition than your $4-\frac{a}{2} \ge 0$.
Finally, as you wrote, the solution is obtained by combining (more precisely, taking the union of) the sets of values of $a$ resulting from each of your $3$ cases.
