Finding variance of the sample mean of a random sample of size n without replacement from finite population of size N.

It is time to upgrade the methods you rely on, since listing every possibility yields the result for small values of $n$ and $N$ but this approach (1) leads to a dead end for general values (as you realized), and (2) provides no insight.

So... let us attack directly (c), considering $N$ chips numbered from $1$ to $N$, and samples of size $n$. Then the sample mean $\bar X$ is such that $n\bar X=\sum\limits_{k=1}^nY_k$ where $Y_k$ is the $k$th chip. By hypothesis, each $Y_k$ is uniform on $\{1,2,\ldots,N\}$ hence $E(Y_k)=\frac1N\sum\limits_{x=1}^Nx=\frac12(N+1)$ for every $k$ and, by linearity of the expectation, $$E(\bar X)=\frac{N+1}2.$$ Likewise, $n^2\bar X^2=\sum\limits_{k=1}^nY_k^2+\sum\limits_{k\ne\ell}Y_kY_\ell$, $E(Y_k^2)$ does not depend on $k$ and $E(Y_kY_\ell)$ does not depend on $k\ne\ell$, hence $nE(\bar X^2)=E(Y_1^2)+(n-1)E(Y_1Y_2)$ hence it suffices to compute $E(Y_1^2)$ and $E(Y_1Y_2)$. By the same argument as before, $E(Y_1^2)=\frac1N\sum\limits_{x=1}^Nx^2=\frac16(N+1)(2N+1)$. To compute $E(Y_1Y_2)$, one can note that, conditionally on $Y_k=x$, $Y_\ell$ is independent on $\{1,2,\ldots,N\}\setminus\{x\}$ and proceed.

Or, one can use the specific case $n=N$, since then, $N\bar X=\sum\limits_{x=1}^Nx$ with full probability, in particular, $NE(Y_1^2)+N(N-1)E(Y_1Y_2)=\frac14N^2(N+1)^2$, which leads to $E(Y_1Y_2)=\frac1{12}(N+1)(3N+2)$. Finally, $$E(\bar X^2)=\frac{2(N+1)(2N+1)+(N+1)(3N+2)(n-1)}{12n},$$ and the variance follows.

Sanity check: If $n=N$, then $E(\bar X^2)=E(\bar X)^2$ (do you see why?).