Question 36, Let $\left.f(x)=a x^{3}+b x^{2}+c x+d （ a \neq 0, a 、 b 、 c 、 d \in R\right)$, given that $0<2 f(2)=$ $3 f(3)=4 f(4)<1$, try to find the range of values for $f(1)+f(5)$.
Question 36, Solution Method One: Let $2f(2) = 3f(3) = 4f(4) = t \in (0,1)$. Notice:
$$
\Delta^{4} f(x+1) = \sum_{k=0}^{4} \binom{4}{k} (-1)^{4-k} f(x+k+1)
$$

and $\Delta^{4} f(x) = 0$, so
$$
\begin{array}{l}
0 = \Delta^{4} f(1) = f(5) - 4f(4) + 6f(3) - 4f(2) + f(1) = f(1) + f(5) - t \\
\Rightarrow f(1) + f(5) = t \\
f(0) + f(4) = t \in (0,1).
\end{array}
$$

Thus, $f(0) + f(4) = t \in (0,1)$.
Solution Method Two: Let $2f(2) = 3f(3) = 4f(4) = t \in (0,1)$.
According to the Lagrange interpolation formula, we have:
$$
f(x) = \sum_{k=1}^{5} \prod_{i \neq k} \frac{(x-1)}{(k-1)} \cdot f(k)
$$

By comparing the coefficients of the fourth power on both sides, we get:
$$
\begin{array}{l}
0 = \sum_{k=1}^{5} \prod_{i \neq k} \frac{f(k)}{(k-1)} = \frac{f(1)}{4!} - \frac{f(2)}{3! \cdot 1!} + \frac{f(3)}{2! \cdot 2!} - \frac{f(4)}{1! \cdot 3!} + \frac{f(5)}{4! \cdot 1!} = \frac{f(1) + f(5) - t}{24} \\
\Rightarrow f(1) + f(5) = t
\end{array}
$$

Thus, $f(1) + f(5) = t \in (0,1)$.