1.
Let $n$ be a positive odd integer.
Let $a, b$ be integers such that $a \equiv b$ (mod $n$).
Then
$$\left(\frac{a}{n}\right) = \left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ with every prime factor repeated according to its multiplicity.
Since $a \equiv b$ (mod $p$) for every prime factor of $n$, $\left(\frac{a}{p}\right) = \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{p}\right) = \prod \left(\frac{b}{p}\right)$.
Hence $\prod \left(\frac{a}{n}\right) = \prod \left(\frac{b}{n}\right)$.
2.
Let $a$ be integers.
Let $m, n$ be positive odd integers.
Then
$$\left(\frac{a}{mn}\right) = \left(\frac{a}{m}\right)\left(\frac{a}{n}\right)$$
Proof:
Clear from the definition of Jacobi symbol.
3.
Let $a, b$ be integers.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$$
Proof:
Let $n = \prod p$ be the prime decomposition of $n$ as in the proof of $1.$
$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$ for every prime factor $p$ of $n$.
Hence $\prod \left(\frac{ab}{p}\right) = \prod \left(\frac{a}{p}\right)\left(\frac{b}{p}\right)$.
Hence $\left(\frac{ab}{n}\right) = \left(\frac{a}{n}\right)\left(\frac{b}{n}\right)$.
Lemma 1
Let $a, b$ be odd integers.
Then $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Proof:
Since $a - 1$ and $b - 1$ are even,
$(a - 1)(b - 1) \equiv 0$ (mod $4$).
Hence $ab - a - b + 1 \equiv 0$ (mod $4$).
Hence $ab - 1 \equiv (a - 1) + (b - 1)$ (mod $4$).
Hence $(ab - 1)/2 \equiv (a - 1)/2 + (b - 1)/2$ (mod $2$).
Lemma 2
Let $a, b$ be odd integers.
Then $(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Proof:
$a^2 - 1 \equiv 0$ (mod $4$).
$b^2 - 1 \equiv 0$ (mod $4$).
Hence
$(a^2 - 1)(b^2 - 1) \equiv 0$ (mod $16$).
Hence
$a^2b^2 - a^2 - b^2 + 1 \equiv 0$ (mod $16$).
Hence $a^2b^2 - 1 \equiv (a^2 - 1) + (b^2 - 1)$ (mod $16$).
Hence
$(a^2b^2 - 1)/8 \equiv (a^2 - 1)/8 + (b^2 - 1)/8$ (mod $2$).
Lemma 3
Let $a, b, c$ be positive odd integers.
Suppose
$\left(\frac{a}{c}\right) \left(\frac{c}{a}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2}}$.
$\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{b-1}{2}\frac{c-1}{2}}$.
Then
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) = (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
Proof:
$\left(\frac{ab}{c}\right) \left(\frac{c}{ab}\right) =\left(\frac{a}{c}\right) \left(\frac{c}{a}\right)\left(\frac{b}{c}\right) \left(\frac{c}{b}\right) = (-1)^{\frac{a-1}{2}\frac{c-1}{2} + \frac{b-1}{2}\frac{c-1}{2}}
= (-1)^{(\frac{a-1}{2} + \frac{b-1}{2})\frac{c-1}{2}}
= (-1)^{\frac{ab-1}{2}\frac{c-1}{2}}$.
The first equality follows from $2.$ and $3.$
The last equality follows from Lemma 1.
4.
Let $m, n$ be positive odd integers such that gcd$(m,n) = 1$.
Then
$$\left(\frac{m}{n}\right) \left(\frac{n}{m}\right) = (-1)^{\frac{m-1}{2}\frac{n-1}{2}}$$.
Proof:
This follows immediately from Quadratic Reciprocity Theorem using Legendre symbol and Lemma 3.
5.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{-1}{n}\right) = (-1)^{\frac{n-1}{2}}$$.
Proof:
This follows immediately from the first supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 1.
6.
Let $n$ be a positive odd integer.
Then
$$\left(\frac{2}{n}\right) = (-1)^{\frac{n^2-1}{8}}$$.
Proof:
This follows immediately from the second supplementary law of quadratic reciprocity using Legendre symbol and $2.$ and Lemma 2.
Best Answer
Suppose that we want to evaluate the Legendre symbol $(a/p)$. If we are going to use Reciprocity, a first step is to factor $a$.
For huge $a$, factorization seems to be computationally very difficult. By doing a Jacobi symbol calculation, we can, apart from trivial divisions by $2$, bypass factorization. The Jacobi symbol calculation gives us an algorithm that works roughly as fast as the Euclidean Algorithm. This makes the calculation feasible for the kinds of numbers involved in current applications.
Remark: The usual version of the Jacobi symbol $(a/m)$ is for odd $m$ only. So your sample calculation is not what I would do. Indeed, one can do similar-flavoured computations and get the wrong answer.
The first step, since $68=2^2\cdot 17$, is to say that $(68/233)=(17/233)$. This is (invert) $(233/17)$, which is $(12/17)$, which is $(3/17)$, which is $(2/3)$, which is $-1$.
I have done quick calculations of the Jacobi symbols in your list. The answers you give are all correct, or at least the same as mine.