$K$ is a simplicial complex of dimension $n$. I am asked to show that $H_i(K)=0$ for all $i>n$.
Why is that something that needs to be shown? Isn't it the simply the defintion of $H_i(K)=0$ for all $i>n$?
I guess that I don't really understand the meaning of $H_i(K)=0$ for $i>n$, so any help in showing that would be appreciated.
I am also asked to show that $H_n(K)$ is a free group, but I have no clue on what should I do. Is it true only for the dimesnion of the simplicial complex? or is it true for other homology groups as well?
Let us remind ourselves of the definition of simplicial homology. Let $C_i(K)$ be the free abelian group generated by the $i$-dimensional simplices, and let $\partial_i : C_i (K) \to C_{i-1}(K)$ be the group homomorphism that sends $i$-dimensional simplices to their $(i-1)$-dimensional boundaries. We have a complex of group homomorphisms, $$ \dots \to C_{i+1}(K) \overset{\partial_{i+1}}{\longrightarrow} C_i(K)\overset{\partial_{i}}{\longrightarrow } C_{i-1}(K) \to \dots$$ Contained within $C_{i}(K)$ are two subgroups: ${\rm Ker}\partial_i$ and ${\rm Im}{\partial_{i+1}}$. The $i$th homology group is defined to be the quotient group, $$ H_i(K) = \frac{{\rm Ker}\partial_i}{{\rm Im}\partial_{i+1}}.$$
Suppose that $K$ is $n$-dimensional. That means that all simplices are of dimension less than or equal to $n$. For $i > n$, $C_{i}(K)$ is the zero group! So the subgroup ${\rm Ker}\partial_i \subset C_{i}(K)$ is also the zero group. What does this tell you about $H_i(K)$ when $i > n$?
Now let's think about $H_n(K)$, where $n$ is the dimension of $K$. Why is this a free group? Well, if $n$ is the dimension of $K$, then $C_{n}(K)$ is non-trivial, but $C_{n+1}(K)$ is trivial, so ${\rm Im}\partial_{n+1}$ is trivial. Hence the $n$th homology group is simply, $$ H_n(K) = {\rm Ker}\partial_n.$$ Your task is to show that this kernel is a free abelian group. Remember, $\partial_n$ is a group homomorphism between $C_n(K)$ and $C_{n-1}(K)$, and these themselves are free abelian groups. To show that $\partial_n$ is free, you may like to use that fact that there exist bases for $C_n(K)$ and $C_{n-1}(K)$ in which the matrix representing $\partial_n$ is in Smith normal form (see here).
You also asked whether $H_i(K)$ must be free for other $i$. The answer is no. I encourage you to work out the homology groups for the Klein bottle, and you will see for yourself!