How to prove by pumping lemma these languages are not regular? 
*

*$L_1 = (a^k * b^r \mid k \neq r^2)$

*$L_2 = (a ^{\sum_i ^n t} \mid n > 0 )$

Assume that $L_1$ is regular. Let $n$ be a constant from pumping lemma (it depends only on language!). Consider word $w=a^{n!}b^{n!}$ which is clearly in $L_1$. By pumping lemma there exists words $w_1$, $w_2$ and $w_3$ such that $w=w_1w_2w_3$, $|w_2|>0$, $|w_1w_2|\leq n$ and $\forall_{i\in\mathbb{N}}w_1w_2^iw_3\in L_1$. So $w_2=a^l$ for some $l\in\mathbb{N}$ (because first $n$ letters of $w$ are all $a$s). But then it is easy to see that we can choose $i$ such that $w_1w_2^iw_3 = a^{(n!)^2}b^{n!} \notin L_1$
Now use a similar argument to prove that $L_2$ is not regular
EDIT: How does pumping lemma works? It says that if language $L$ is regular then there exists magic constant $n$ such that we can split every word $w\in L$ longer than $n$, into three parts $w_1w_2w_3$ (with $|w_1w_2|\ge n$) and "pump" the middle term as much as we want always obtaining a word from $L$. So in order to prove that language is not regular all we have to do is to find a word which lacks this property.
For $L_1$ if we take $w=a^{n!}b^{n!}$ and split it into $w_1w_2w_3$, there must be $w_1=a^k$, $w_2=a^l$ and $w_3=a^{n!-k-l}b^{n!}$ (not just $b^{n!}$ as you posted in your comment) with $k+l\le n$. So how does $w_1w_2^iw_3$ look like? It's $$a^ka^{li}a^{n!-k-l}b^{n!}=a^{n!+(i-1)l}b^{n!}$$Can we find $i$ such that $a^{n!+(i-1)l}b^{n!}\notin L_1$? Yes, thanks to factorial. We have $l\le n$, so both $(n!)^2$ and $n!$ are divisible by $l$. We only need to solve equation $n!+(i-1)l=(n!)^2$ which gives us $$i=\frac{(n!)^2-n!}{l}+1$$ and it is indeed a natural number. This contradicts pumping lemma and shows that $L_1$ is not regular.
Different (simpler) approach:
We can use a fact that language $L$ is regular iff it's complement $\bar{L}$ is regular. So it's sufficient to show that $\bar{L_1}=(a^kb^r|k=r^2)$ is not regular. So assume that $\bar{L_1}$ is regular and let $n$ be a constant from pumping lemma. Now looking at word $w=a^{n^2}b^n$ we see that if we split it into $w=w_1w_2w_3$ with $|w_1w_2|\le n$ then $w_2=a^k$ for some $k>0$ and $w_1w_2^2w_3=a^{n^2+k}b^n$ which is not in $\bar{L_1}$ because $n^2+k \neq n^2$
