Use L'Hopital's Rule to Prove Let $$f: \mathbb R\rightarrow \mathbb R$$ be differentiable, let a in $\mathbb R$. Suppose that $f''(a)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}=f''(a) $$

Suppose further that $f''(x)$ exists for all $x$, and that $f'''(0)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=f'''(0)$$
Question 1
$$
\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} \\
$$
when $h = 0$ is substituted numerator and denominator reduce to $0$. So, applying L'Hopital's rule (differentiate wrt $h$)
$$
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)-f'(a-h)+f'(a)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)}{2h}+\lim_{h\rightarrow0}\frac{f'(a)-f'(a-h)}{2h} \\
= \frac{2f''(a)}{2} = f''(a)\\
$$
Question 2
$$
\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}
$$
when h=0 is substituted numerator and denominator reduce to 0. So, applying L'Hopital's rule (differentiate wrt h)
$$
\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \frac{2}{3}\left(\lim_{h\rightarrow0}\frac{f''(h)-f''(0)}{h}+\lim_{h\rightarrow0}\frac{f''(0)-f''(-h)}{h}\right)+\frac{1}{3}\left(\lim_{h\rightarrow0}\frac{f''(0)-f''(h/2)}{h}+\lim_{h\rightarrow0}\frac{f''(-h/2)-f''(0)}{h}\right) \\
= \frac{4f'''(0)}{3}-\frac{f'''(0)}{3} = f'''(0)
$$
PS: Thanks to @Paramanand Singh for pointing out that f was not twice and thrice differentiable in first and second questions respectively.
The first proof:
$$lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}=$$using De L'Hopital Rule
$$=lim_{h\rightarrow0}\frac{f'(a+h)+f'(a-h)}{2h}=\frac{f''(a+h)+f''(a-h)}{2}=\frac{f''(a)+f''(a)}{2}=f''(a)$$
Instead the second question:
$$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=$$ applying De L'Hopital Rule
$$\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(-h)-\frac{1}{2}f''(h/2)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f'''(h)+f'''(-h)-\frac{1}{4}f'''(h/2)-\frac{1}{4}f'''(-h/2))}{6} \\
= \frac{6f'''(0)}{6} = f'''(0)$$