How to find the limit of series? (What should I know?)

Hint:

All you have to know is the sum of the $n$ first terms of a geometric series, which is a formula from high school: $$1+q+q^2+\dots +q^n=\frac{1-q^{n+1}}{1-q}\qquad (q\ne 1),$$from which we can deduce: $$q^r+q^{r+1}+\dots +q^n=q^r(1+q+\dots+q^{n-r})=q^r\frac{1-q^{n-r+1}}{1-q}=\frac{q^r-q^{n+1}}{1-q}.$$ If $\lvert q\rvert<1$, these sums have limits $\;\dfrac1{1-q}\;$ and $\;\dfrac{q^r}{1-q}\;$ respectively.

What should I do when I need a limit of infinite sum? (are there any rules of thumb?)

In general - no there aren't such rules. There are specific types of series for which it is known how to compute the limit (like the geometric series). There are way more recepies for figuring out whether a series converges or not (finding the limit is much harder). But also here the cookbook is limited. As others pointed out already, your two series are geometric series, who's limit can be computed easily.

$$\color{red}{1+}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \ldots + \frac {1}{2^{n}}$$ Is an example of $$1+x+x^2+x^3+\ldots+x^n$$ Where the ratio between a number $a_n$ and the previous $a_{n-1}$ is constant and is $x$. This is the sum of a geometric progression and it's quite easy to see that its value is $$\frac{1-x^{n+1}}{1-x}$$ When $n\to\infty$ the sum converges only if $|x|<1$ because if so $x^{n+1}\to 0$ and the sum is $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$ In your first example $x=\frac12$ and index $n$ starts from $n=1$ so the sum is $$\sum_{n=1}^{\infty}\left(\frac12\right)^n=\frac{1}{1-\frac12}-1=\color{red}{1}$$ The second one is $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n}}{3^{n}}=\sum_{n=0}^{\infty}\left(-\frac13\right)^n=\frac{1}{1-\left(-\frac13\right)}=\color{red}{\frac34}$$

Hope this is useful