Are these the notes you took in class? ie.. do you just want some explanation on some of the details? Well for your first example, in the last step the $$1\over 5$$ is multiplied into the summation giving the term $$x^{n+1}\over {5^{n+1}(n+1)}$$. Then the power series is shifted by replacing n with n-1 (which explains the term you have written in the last step). However, in doing this, the series now starts at $n=1$ not $n=0$. Was that a typo? Clearly the term to the right of the last summation is undefined at $n=0$
For the second example (and I guess the first one too) the term $C$ is an arbitrary constant of integration. So to evaluate the definite integral in your second question, you can choose $C$ to be zero as any antiderivative can be used in the fundamental theorem of calculus.
In general, power series are nice because polynomials are by far the easiest functions to work with, and a power series allows you approximate a much harder function as a polynomial. There are only a few well known power series in elementary calculus (namely, $sin(x)$, $cos(x)$, $e^x$, and $1\over {x-1}$ ) but by differentiating and integrating you can produce power series for many, many functions. That's the whole point of this.
This looks fine! Just a bit of minor nitpicking:
1) Looks good!
2) Looks good!
3) Your formula for the remainder term looks off. It should be $R_n(x) = \frac{(x - x_0)^{n + 1}}{(n+1)!} f^{(n + 1)}(\xi)$ [wikipedia]
Edit: as OP mentioned, their version reduces to this version when $p=n+1$. I couldn't find any references to validate your version though, but as long as it comes from a credible source it should be fine.
4) Remember, $\lim\limits_{n\to\infty} R_n(x)=0$ needs to hold for all $x$
5) Really, this is the same as step 4 -- if the error term tends to $0$ then the functions are the same. So step $5$ isn't even necessary.
Edit: This is only the same as step 4 if for step 4 you proved that the error converges to $0$ for all $x\in\mathbb{R}$, not just for a single interval. Of course, if you can prove that it converges to $0$ for any interval then that'd work too. Otherwise, if you only did step 4 for a specific interval, it tells you nothing about what happens outside that interval.
6) If we followed the steps correctly, and all conditions were met,
$$f(x)=\sum_{n = 0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n$$
is indeed the Taylor (power) series around $x_0$, it converges for all $x$ to $f(x)$, and $f$ is indeed analytic.
But, there's an important point I want to make -- some functions (e.g $\log$) are analytic without a corresponding Taylor series equal to the function. Your argument can only find the Taylor series for a function, and prove that it's analytic, but there are analytic functions which you might miss if you use this method.
Your understanding on this topic is very good :) the best thing to do now is to do some examples. I would recommend trying $x^2$ as a warm up, or $e^x$ if you're feeling more confident.
Best Answer
Hint
You started well using $$ln(1+x)=\sum_{n=0}^\infty {(-1)^{n+1}\frac{x^n}{n}}$$ Now change $x$ to $-x$ to get $$ln(1-x)=-\sum_{n=0}^\infty {\frac{x^n}{n}}$$ Now replace $x$ by $t^2$ and finish.
I am sure you can take from here.