Part 1
First let’s write out the Taylor expansion for 2f(x+h) and f(x-2h).
f(x-2h) \approx f(x) - 2f'(x)h + 2f''(x)h^{2} + O(h^{3})
2f(x+h) \approx 2f(x) + 2f'(x)h + f''(x)h^2 + O(h^3)
If we add these two equations together we get:
f(x-2h) - 3f(x) + 2f(x+h) \approx 3f''(x)h^2 + 2O(h^3).
Now divide both sides by 3h^2.
f''(x) \approx \frac{2f(x+h) - 3f(x) + f(x-2h)}{3h^2} - O(h)
Part 2
Our value of h will be \Delta x since we are restricted by these four data points. So, h=0.1. The question does not specify which x value that we need to approximate the second derivative for, but as it turns out, f''(0.2) is the only one allowed given the data.
f''(0.2) \approx \frac{2f(0.3) - 3f(0.2) + f(0)}{0.03}
\approx \frac{2(1.27) - 3(1.22) + 1}{0.03}
f''(0.2) \approx -4
