Here's a lot more help than your question warrants...
a) one-to-one means if f(a) = f(b), then a = b
i.e. if (a - 2)^5 = (b - 2)^5, then a = b.
This is true after taking (fifth) roots.
b). Look up partial fractions. Or just have wolfram alpha do it...
2) What you have to do is the following:
You have to find $\displaystyle A,B \ \mbox{and} \ C$ in the expression:
$\displaystyle \frac{3x+11}{(x-3)(x^2+1)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+1}$
So make a common denominator in the RHS, like this:
$\displaystyle \frac{A}{x-3}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)\(x-3)}{(x-3)(x^2+1)}=...$
Go further (expand the numerator), afterwards you have to make a (small) system to determine A,B and C.
If you expand the numerator of the RHS then you get:
$\displaystyle \frac{3x+11}{(x-3)(x^2+1)}=\frac{A(x^2+1)+(Bx+C)(x-3)}{(x-3)(x^2+1)}\\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{Ax^2+A+Bx^2+Cx-3Bx-3C}{(x-3)(x^2+1)}\\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{(A+B)x^2+(C-3B)x+(A-3C)}{(x-3)(x^2+1)}$
You notice the denominator's are equal so the numerator's have to be equal therefore:
$\displaystyle 3x+11=(A+B)x^2+(C-3B)x+(A-3C)$
And so:
$\displaystyle C-3B=3$, $\displaystyle A-3C=11$ and $\displaystyle A+B=0$ (why?)
Solve this simultaneous equations.
Or, at this point, recognizing that, since the denominators are the same, the numerators must be the same,
$\displaystyle 3x+ 11= A(x^2+1)+ (Bx+ C)(x- 3)$
Taking x= 3, that gives 3(3)+ 11= 20= A(10) so that A= __.
Taking x= 0, 3(0)+11= 11= A+ C(-3)= 2- 3C so that C= __.
Taking x= 1, 3(1)+ 11= 14= A(2)+(B+ C)(-2)= 4+ B- 3 so B= __
$\displaystyle \\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{Ax^2+A+Bx^2+Cx-3Bx-3C}{(x-3)(x^2+1)}\\ \\ \\ \phantom{xxxxxxxxxxxxxxx}=\frac{(A+B)x^2+(C-3B)x+(A-3C)}{(x-3)(x^2+1)}$
You notice the denominator's are equal so the numerator's have to be equal therefore:
$\displaystyle 3x+11=(A+B)x^2+(C-3B)x+(A-3C)$
And so:
$\displaystyle C-3B=3$, $\displaystyle A-3C=11$ and $\displaystyle A+B=0$ (why?)
Solve this simultaneous equations.