In general, the Maclaurin series of $f$ is: (which is the Taylor series of $f$ at $a=0$)
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n.$$
Therefore, the first four terms of the Maclaurin series of $f$ are:
$$\tag{1}\sum_{n=0}^3\frac{f^{(n)}(0)}{n!}x^n=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3.$$
In your case, $f(x)=x^3 - 2x^2 + 2x - 3$. Then you can calculate $f(0)$, $f'(0)$, $f''(0)$, and $f'''(0)$ and put them back into $(1)$ to get the answer. For example,
$$f(0)=0^3 - 2\cdot0^2 + 2\cdot 0- 3=-3, f'(0)=(3x^2-4x+2)\big|_{x=0}=3\cdot 0^2-4\cdot 0+2=2.$$ I will let you do $f''(0)$ and $f'''(0)$.
The Maclaurian series of $f$ (Taylor series of $f$ at $0$) with one term is $f(0)=-3$.
The Maclaurian series of $f$ with two terms is $f(0)+\displaystyle\frac{f'(0)}{1!}x=-3+2x$.
The Maclaurian series of $f$ with three terms is $f(0)+\displaystyle\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2=-3+2x-2$
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This will be rather difficult.
Let's first of all suppose there are no problems, i.e. the functions you are interested in are infinitely differentiable and continuous in the range of values you are interested in.
Then, even with the simple example that you know the Maclaurin series of $f(x)$ and want the Taylor series of $f(x)$ about another point $a$, this is a cumbersome endeavour. Here is the procedure.
With the known Maclaurin series (you memorized the $m_k$)
$f(x) = \sum\limits_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} x^k = \sum\limits_{k = 0}^{\infty} m_k x^k$
you require the Taylor series
$f_a(x) = \sum\limits_{k = 0}^{\infty} c_k (x-a)^k$
Now the coefficients are given by
$c_k = \frac{f^{(k)} (a)}{k!}$
which in turn you can get by differentiating your MacLaurin series and evaluate at $a$:
$c_k = \frac{f^{(k)} (a)}{k!} = \sum\limits_{n = k}^{\infty} \frac{n!}{k! (n-k)!} m_n a^{n-k}$
E.g.
$c_0 = \sum\limits_{n = 0}^{\infty} m_n a^{n}$
So you need to sum an infinite series to arrive at your new coefficients - big labour.
Another method is to identify the coefficients of all powers of $(x-a)$. To do so, write the MacLaurin series
$f(x) = \sum\limits_{k = 0}^{\infty} m_k ((x-a)+a)^k
= \sum\limits_{k = 0}^{\infty} m_k \sum_{j=0}^k {k \choose j} (x-a)^j (a)^{k-j}$
This leads to the same result.
Things get worse if you consider $g(f(x))$.
I guess this already answers your question: it's not a smooth technique to calculate the Taylor expansion from the known MacLaurin series.
Best Answer
Let us first deal with the function itself. The MacLaurin expansion has the shape $1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots$. Integrating from $0$ to $x$, we find that the Maclaurin expansion for the area up to $x$ has Maclaurin expansion $x-\frac{x^3}{3\cdot 1!}+\frac{x^5}{5\cdot 2!}-\frac{x^7}{7\cdot 3!}+\cdots$.
For the function, you were asked to truncate just after the $\frac{x^2}{1!}$ term. You can estimate the error in terms of the third derivative, evaluated at some unknown place $\xi$ between, in this case, $0$ and $0.5$. Now in fact this $\xi$ is close to $0$, but you do not officially know that, and the upper bound on the error that you get is quite a bit too pessimistic.
However, as you observed, the coefficient of $x^3$ is $0$, and therefore the Maclaurin polynomial up to $x^2$ is exactly the same as the Maclaurin polynomial up to $x^3$. So the error can be expressed in terms of the fourth derivative, and this gives a nicer upper bound on the error.
There is another way of looking at things that does not use the Lagrange formula for the remainder. Note that the Maclaurin polynomial, at least for $x\le 1$, is an alternating series. So the truncation error is less, in absolute value, than the first "neglected" term. This term has absolute value $\frac{x^4}{2!}$. So the absolute value of the error when we evaluate at $x=0.5$ is less than $\frac{(0.5)^2}{2!}$.
The error estimation for the integral uses exactly the same ideas.