To find the constants in the rational fraction $\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$
you may use any set of 3 values of $x$, provided that the denominator $x^3-4x^2+4x\ne 0$.
The "standard" method is to compare the coefficients of $\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$ after multiplying this rational fraction by the denominator $x^3-4x^2+4x=x(x-2)^2$ and solve the resulting linear system in $A,B,C$. Since
$\begin{eqnarray*} \frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} &=&\frac{A}{x}+\frac{B}{x-2}+% \frac{C}{\left( x-2\right) ^{2}} \\ &=&\frac{A\left( x-2\right) ^{2}}{x\left( x-2\right) ^{2}}+\frac{Bx\left( x-2\right) }{x\left( x-2\right) ^{2}}+\frac{Cx}{x\left( x-2\right) ^{2}} \\ &=&\frac{A\left( x-2\right) ^{2}+Bx\left( x-2\right) +Cx}{x\left( x-2\right) ^{2}} \\ &=&\frac{\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A}{x\left( x-2\right) ^{2}}, \end{eqnarray*}$
if we equate the coefficients of the plynomials
$x^{2}+3x-4\equiv\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A,$
we have the system
$\begin{eqnarray*} A+B &=&1 \\ -4A-2B+C &=&3 \\ 4A &=&-4, \end{eqnarray*}$
whose solution is
$\begin{eqnarray*} B &=&2 \\ C &=&3 \\ A &=&-1. \end{eqnarray*}$
Alternatively you could use the method indicated in parts A and B, as an example.
A. We can multiply $f(x)$ by $x=x-0$ and $\left( x-2\right) ^{2}$ and let $% x\rightarrow 0$ and $x\rightarrow 2$. Since $\begin{eqnarray*} f(x) &=&\frac{P(x)}{Q(x)}=\frac{x^{2}+3x-4}{x^{3}-4x^{2}+4x} \\ &=&\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} \\ &=&\frac{A}{x}+\frac{B}{x-2}+\frac{C}{\left( x-2\right) ^{2}},\qquad (\ast ) \end{eqnarray*}$ if we multiply $f(x)$ by $x$ and let $x\rightarrow 0$, we find $A$: $A=\lim_{x\rightarrow 0}xf(x)=\lim_{x\rightarrow 0}\frac{x^{2}+3x-4}{\left( x-2\right) ^{2}}=\frac{-4}{4}=-1.$ And we find $C$, if we multiply $f(x)$ by $\left( x-2\right) ^{2}$ and let $x\rightarrow 2$: $C=\lim_{x\rightarrow 2}\left( x-2\right) ^{2}f(x)=\lim_{x\rightarrow 2}\frac{% x^{2}+3x-4}{x}=\frac{2^{2}+6-4}{2}=3.$ B. Now observing that $P(x)=x^{2}+3x-4=\left( x+4\right) \left( x-1\right)$ we can find $B$ by making $x=1$ and evaluate $f(1)$ in both sides of $(\ast)$, with $A=-1,C=3$: $f(1)=0=2-B.$ So $B=2$. Or we could make $x=-4$ in $(\ast)$ $f(-4)=0=\frac{1}{3}-\frac{1}{6}B.$ We do obtain $B=2$.
Thus $\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}}=-\frac{1}{x}+\frac{2}{x-2}+\frac{3% }{\left( x-2\right) ^{2}}\qquad (\ast \ast )$
Remark: If the denominator has complex roots, then an expansion as above is not possible. For instance
$\frac{x+2}{x^{3}-1}=\frac{x+2}{(x-1)(x^{2}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{% x^{2}+x+1}.$
You should find $A=1,B=C=-1$:
$\frac{x+2}{x^{3}-1}=\frac{1}{x-1}-\frac{x+1}{x^{2}+x+1}.$