Let $A$ be a square $n \times n$ matrix, which is invertible. Now we want to find the $i$th column of $A^{1}$ and one $(i,j)$ entry of $A^{1}$. Is there any way to compute only a small of portion of the entries in $A^{1}$?

The inverse of a matrix is the adjoint divided by the determinant. So what you want to compute is the determinant of an $(n1) \times (n1)$ submatrix, divided by the determinant of your original matrix. Asymptotically, this won't be much saving, if any, over just inverting $A$, but it might be easier to think about, or might be simpler for matrices with a special form. Also, if you care about efficiency, remember that you should never compute a determinant, or an inverse, by summing over all $n!$ permutations! Use row reduction or, better, use a good linear algebra library. 


If you are willing to settle for an approximate answer, then have a look at the paper: Z. Bai, M. Fahey and G. Golub, Some largescale matrix computation problems Journal of Computational and Applied Mathematics, Volume 74, Issues 12, 5 November 1996, Pages 7189. Link to PDF That paper studies simple randomized methods for estimating $$u^Tf(X)v,$$ where $f(X)$ is a matrix function. In your case, $f(X) = X^{1}$. If you choose $u=e_i$ and $v=e_j$ (the standard basis vectors), then you can extract an approximation to $(X^{1})_{ij}$. In the meanwhile there is more related work, but if you are happy with these types of results that I can expand my answer to provide some more citations. 


The ith column of $A^{1}$ is $A^{1}$ applied to the ith unit vector. So you can obtain it by solving a linear system. This question does not appear suitable for this site. 


To amplify on @Michael Renardi's answer: finding $x$ such that $A x = y$ (in this case, $y$ is a unit vector) is generally much faster than finding $A^{1},$ the key words are "conjugate gradient algorithm. I do not understand the question about the $(i, j)$th entry, since once you fiund the column (or the row, by taking transposes) you can find your element. 

