Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively.

Hint

You have $$f(x)=(x^2-1)(ax+b)+(2x-5)$$ and $$f(x)=(x^2-4)(cx+d)+(-3x+4)$$ From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus $$\color{red}{c+d=\frac{4}{3}}$$

From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus $$\color{red}{2a+b=\frac{-1}{3}}$$ You also get $$f(0)=-b-5=-4d+4 \implies\color{red}{4d-b=9}$$

NOTE You can make a simple observation about the coefficients of $x^3$ and $x^2$ in both the expressions for $f(x)$ to conclude that $a=c$ and $b=d$. With that the first two equations can suffice.

Hint:

Let $$\dfrac{f(x)}{(x-1)(x+1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{x-2}+\dfrac D{x+2}$$

$$\implies f(x)=?$$

where $A,B,C,D$ which are arbitrary constants which can be found by putting $x=1,-1,2,-2$ one by one.

For example, put $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$

You have (for instance) that $f(x) = Q(x)(x+1)(x-1) + R(x)$, where $R(x) = 2x-5$.

Putting $x=1$ then $x=-1$ gives you $f(1) = - 3$ and $f(-1)=-7$.

Doing the same with $x=2, x=- 2$ allows you to work out two more points on the cubic.

Four distinct points allow you to completely characterise a cubic, you have a system of four linear equations to solve for the coefficients.