The nodes of the $n$-point Gaussian quadrature are given by the zeros of the orthogonal polynomial $\phi_n(x)$. For your case, that would be the roots of
$$
x^2 -4x + 2 = 0\\
x_{1,2} = 2 \pm \sqrt{2}.
$$
Weights of an interpolating quadrature
Consider some arbitrary interpolating type quadrature with nodes $x_i$ and weights $w_i$. A quadrature is said to be interpolating if it exactly integrates an interpolating polynomial $P(x)$ passing through points $(x_i, f(x_i))$. Polynomial $P(x)$ could be expressed in terms of Lagrange interpolation basis
$$
f(x) \sim P(x) = \sum_{i=1}^n f(x_i) \ell_i(x), \qquad \ell_i(x) = \frac{\prod_{j\neq i}(x-x_j)}{\prod_{j\neq i}(x_i-x_j)}.
$$
Integrating the both sides gives a rule
$$
\int_a^b f(x)\omega(x)dx \approx
\int_a^b P(x)\omega(x)dx =
\sum_i f(x_i) w_i\\
w_i \equiv \int_a^b \ell_i(x) \omega(x) dx.
$$
So, for an interpolating type quadrature (which Gaussian quadratures are) the weights can be computed by integrating Lagrange basis functions
$$
w_i = \int_a^b \ell_i(x) \omega(x) dx.
$$
This formula would integrate every $f(x)$ precisely, provided that $f(x)$ is a polynomial of degree $n-1$, since $P(x)$ would just be $f(x)$ itself. Note that this holds whatever nodes $x_i$ we choose. So careful choice of $x_i$ may (and does) extend this property for higher polynomial degrees.
Relation between Gaussian quadratures and orthogonal polynomials
Consider some polynomial $f(x)$ with degree $m \geq n$. We can divide it by $(x-x_1)\cdots(x-x_n)$ polynomial
$$
f(x) = q(x)(x-x_1)\cdots(x-x_n) + r(x),\qquad \operatorname{deg}(q) = m - n,\quad \operatorname{deg}(r) \leq n-1.
$$
Applying quadrature rule to $f(x)$ yields
$$
\int_a^b f(x) \omega(x) dx \approx \sum_{i=1}^n w_i f(x_i) =
\sum_{i=1}^n w_i r(x_i).
$$
Quadrature rule ignores $q(x)(x-x_1)\cdots(x-x_n)$ part since it vanishes as every node $x_i$. We want the same property for the integral
$$
\int_a^b q(x)(x-x_1)\cdots(x-x_n)\omega(x) dx = 0
$$
for every polynomial $q(x)$ with its degree not exceeding some upper limit. Consider case when $x_i$ are roots of $\phi_n(x)$ orthogonal polynomial. Then
$$
\phi_n(x) = A_n(x-x_1)\cdots(x-x_n)
$$
and $q(x)$ can be expanded over basis of $\phi_0(x), \phi_1(x), \dots$:
$$
q(x) = q_0 \phi_0(x) + q_1 \phi_1(x) + \dots
$$
Now
$$
\int_a^b q(x)(x-x_1)\cdots(x-x_n)\omega(x) dx =
\frac{1}{C_n} \int_a^b q(x)\phi_n(x) \omega(x) dx =
\frac{1}{C_n} \sum_{k > 0} q_k \langle \phi_k, \phi_n \rangle
= \frac{q_n\gamma_n}{A_n}.
$$
That integral would be zero if $q_n = 0$ which is true for every polynomial of degree less than $n$, i.e. $\operatorname{deg}(q) < n$ and
$\operatorname{deg}(f) < n+n = 2n$.
Expressing weights without Lagrange basis functions
Observe that
$$
\ell_i(x) = \frac{\prod_{j \neq i}(x - x_j)}{\prod_{j \neq i}(x_i - x_j)} = \frac{\phi_n(x)}{A_n(x-x_i)\prod_{j \neq i}(x_i - x_j)}.
$$
Let's consider the $\phi_n'(x)$:
$$
\phi_n'(x) = A_n\frac{d}{dx}\prod_{i=1}^n (x - x_i) =
A_n\sum_{j=1}^n \prod_{i \neq j} (x - x_i).
$$
At nodes only one term lefts from the sum:
$$
\phi'_n(x_k) = A_n\sum_{j=1}^n \prod_{i \neq j} (x_k - x_i) =
\prod_{i \neq k} (x_k - x_i).
$$
Thus
$$
\ell_i(x) = \frac{\phi_n(x)}{(x-x_i)\phi'_n(x_i)}\\
\int_a^b \ell_i(x) \omega(x) dx =
\frac{1}{\phi'(x_i)}\int_a^b \frac{\phi_n(x)}{x-x_i} \omega(x) dx.
$$
Now let's evaluate $\int_a^b \frac{\phi_n(x)}{x-x_i} \omega(x) dx$ using auxillary integral
$$
\int_a^b \frac{\phi_n(x)}{x-x_i} [\phi_{n+1}(x)-\phi_{n+1}(x_i)] \omega(x) dx =
\int_a^b \phi_n(x)\frac{\phi_{n+1}(x)-\phi_{n+1}(x_i)}{x-x_i} \omega(x) dx.
$$
Since $\phi_{n+1}(x)-\phi_{n+1}(x_i)$ vanishes at $x = x_i$ by polynomial remainder theorem $\frac{\phi_{n+1}(x)-\phi_{n+1}(x_i)}{x-x_i}$ is a polynomial of degree $n$ with leading coefficient $A_{n+1}$.
$$
\frac{\phi_{n+1}(x)-\phi_{n+1}(x_i)}{x-x_i} = \frac{A_{n+1}}{A_n} \phi_n(x) + \text{polynomial rest of degree } < n.
$$
The polynomial $\phi_n(x)$ is orthogonal to the polynomial rest, auxiliary integral is
$$
\int_a^b \frac{\phi_n(x)}{x-x_i} [\phi_{n+1}(x)-\phi_{n+1}(x_i)] \omega(x) dx =
\frac{A_{n+1}}{A_n} \int_a^b \phi_n^2(x) dx = \frac{A_{n+1}}{A_n}\gamma_n\\
\int_a^b \frac{\phi_n(x)}{x-x_i} \phi_{n+1}(x) \omega(x) dx
-\phi_{n+1}(x_i)\int_a^b \frac{\phi_n(x)}{x-x_i} \omega(x) dx =
\frac{A_{n+1}}{A_n}\gamma_n.
$$
On the other hand the integral
$$
\int_a^b \frac{\phi_n(x)}{x-x_i} \phi_{n+1}(x) \omega(x) dx
$$
is zero, because $\frac{\phi_n(x)}{x-x_i}$ is a polynomial of degree less than $n+1$.
Finally
$$
\int_a^b \frac{\phi_n(x)}{x-x_i} \omega(x) dx =
-\frac{A_{n+1}\gamma_n}{A_n\phi_{n+1}(x_i)}
$$
and
$$
w_i = -\frac{A_{n+1}\gamma_n}{A_n\phi_{n+1}(x_i)\phi'_n(x_i)}.
$$
This is the correct version of the first formula from (c). It can be easily seen that the formula should stay invariant under scaling of the polynomials $\phi_i(x)$, i.e. changing $A_k$ (note only that $\gamma_k \sim A_k^2$):
$$
w_i \sim \frac{A_{n+1}A_n^2}{A_n A_{n+1} A_n} = 1.
$$
Relation between recurrence coefficients $a_n, c_n$ and $A_k, \gamma_k$
We're given that
$$
\phi_{n+1}(x) - (a_n x +b_n) \phi_n(x) + c_n \phi_{n-1}(x) = 0.
$$
Plugging $x = x_i$ zeroes $\phi_n(x)$ so
$$
\phi_{n+1}(x_i) = -c_n \phi_{n-1}(x_i).
$$
Now we have to express $c_n$ using known values of $A_n$ and $\gamma_n$.
$$
0 = \langle
\phi_{n+1}(x) - (a_n x +b_n) \phi_n(x) + c_n \phi_{n-1}(x)\;,\;
\phi_{n+1}(x)
\rangle
= \gamma_{n+1} - a_n \langle x \phi_n(x) , \phi_{n+1}(x) \rangle\\
0 = \langle
\phi_{n+1}(x) - (a_n x +b_n) \phi_n(x) + c_n \phi_{n-1}(x)\;,\;
\phi_{n-1}(x)
\rangle
= c_n\gamma_{n-1} - a_n \langle x \phi_n(x) , \phi_{n-1}(x) \rangle\\
$$
Since $\langle x \phi_n(x) , \phi_{n-1}(x) \rangle = \langle \phi_n(x) , x\phi_{n-1}(x) \rangle$ we have
$$
\gamma_{n+1} = a_n \langle x\phi_n , \phi_{n+1} \rangle\\
c_n\gamma_{n-1} = a_n \langle \phi_n, x \phi_{n-1} \rangle\\
\frac{\gamma_n}{a_{n-1}}= \langle \phi_n, x \phi_{n-1} \rangle = \frac{c_n \gamma_{n-1}}{a_n}
$$
so
$$
c_n = \frac{a_n}{a_{n-1}} \frac{\gamma_n}{\gamma_{n-1}}.
$$
Notice that $A_{n+1} - a_n A_n$ is the coefficient of the $x^{n+1}$ term in recurrence, so
$$
a_n = \frac{A_{n+1}}{A_n}.
$$
Finally,
$$
c_n = \frac{A_{n+1}A_{n-1}}{A_n^2} \frac{\gamma_n}{\gamma_{n-1}}\\
w_i = \frac{A_{n+1}\gamma_n}{A_n c_n \phi_{n-1}(x_i)\phi'_n(x_i)} =
\frac{A_{n+1}A_n^2\gamma_n\gamma_{n-1}}{A_n A_{n+1}A_{n-1}\gamma_n \phi_{n-1}(x_i)\phi'_n(x_i)} = \frac{A_n\gamma_{n-1}}{A_{n-1} \phi_{n-1}(x_i)\phi'_n(x_i)}.
$$
