Firstly lets write a, b, c and d as:
$$a=x-1$$
$$b=x$$
$$c=x+1$$
$$d=x+2$$
Question 1:
We can say that
$$(a^2+d^2)-(b^2+c^2) = ((x-1)^2 + (x+2)^2) - (x^2 + (x+1)^2)$$
$$=(2x^2 + 2x +5) - (2x^2 +2x +1) = 4$$
which is independent of $x$
Question 2:
We can say that
$$(a^2+b^2 + c^2+d^2) - (1+2+3)= (x-1)^2 + x^2 + (x+1)^2 + (x+2)^2 -6$$
$$=4x^2 + 4x = 4(x^2 + x)$$
so is divisible by 4
Now looking at the bit in the brackets, namely $x^2+x$ it is one of the two following cases:
$$= odd + odd = even$$
$$= even + even = even$$
So is always divisible by 2 making the whole number divisible by 8
I will now see if this is true for the generalized problem by conjecturing that 'The sum of the square of n consecutive integers is divisible by n', namely:
$$\sum_{r=1}^{n} (a_r)^2 - \sum_{r=1}^{n-1} r$$ is divisible by $n$ $\{ a_r$ are consecutive integers $\}$.
I will split this into 2 cases, $n=2k-1$ and $n=2k$ $\{ k \in \mathbb{Z}^+ \}$
Case 1: $n=2k-1$
$$\sum_{r=1}^{2k-1} (x-k+r)^2 - \sum_{r=1}^{2k-2} r =
\sum_{r=1}^{2k-1} \left[ x^2 + 2(r-k)x +(r-k)^2 \right] - \sum_{r=1}^{2k-2} r$$
$$=(2k-1)x^2 + [2k(2k-1) -2k(2k-1)]x + \frac{(k-1)k(2k-1)}{3} - \frac{(2k-2)(2k-1)}{2}$$
$$=(2k-1)x^2 + \frac{k(2k-1)(k-1)}{3}-(k-1)(2k-1)$$
Now we can divide through by $(2k-1)$ we can check whether the above expression is divisible by $(2k-1)$.
$$x^2 + \frac{k(k-1)}{3} - (k-1)$$
The above is an integer when either $k$ is a multiple of 3 or 1 more than a multiple of 3 but NOT when k is 2 more than a multiple of 3 (which does disprove the conjecture for all $n$)
Case 2: $n=2k$
$$\sum_{r=1}^{2k} (x-k+r)^2 - \sum_{r=1}^{2k-1} r$$
(We can use the answer to the last case to help evaluate this as:)
$$=2kx^2 + 2kx + \frac{k(2k-1)(k-1)}{3} + k^2 -(k-1)(2k-1)+(2k-1)$$
$$=2kx^2 + 2kx + \frac{k(2k^2+1)}{3} -(2k-1)k$$
Now we can divide through by $2k$ we can check whether the above expression is divisible by $2k$.
$$x^2 + x + \frac{2k^2+1}{6} -\frac{2k-1}{2} = x^2 + x + \frac{(k-2)(k-1)}{3}$$
The above is an integer when either $k$ is 1 more than a multiple of 3 or 2 more than a multiple of 3 but NOT when k is a multiple of 3.
In conclusion the conjecture is true when $n$ is 1 more than a multiple of 6, 2 more than a multiple of 6, 4 more than a multiple of 6 and 5 more than a multiple of 6 but this is equivalent to saying $n$ is not a multiple of 3 so:
$$n \neq 3k \iff n\mid \left( \sum_{r=1}^{n} (a_r)^2 - \sum_{r=1}^{n-1} r \right) \{ k \in \mathbb{Z}^+ \}$$
Question 3:
Returning back to the original problem of 4 consecutive integers lets explore
$$(x-1)x(x+1)(x+2)$$
Without looking at the algebra we can explain why this must be divisible by 24.
In a set of 4 consecutive numbers at least one is a multiple of 2 a multiple of 3 and a multiple of 4 so:
$$2 \times 3 \times 4 = 24$$
Question 4:
We can say that
$$\sqrt{abcd+1} = \sqrt{(x-1)x(x+1)(x+2) + 1}$$
$$=\sqrt{x^4 + 2x^3 -x^2 -2x + 1} = \sqrt{(x^2 +x-1)^2} = \sqrt{(x+\phi)^2 (x-(\phi - 1))^2}$$
$$=(x+\phi)(x-(\phi - 1))$$
where $\phi$ is the golden ratio
Side note: If we had used the consecutive integers as $(x-2),(x-1),x,(x+1)$ then we would get that
$$\sqrt{abcd+1} = \sqrt{(x^2 -x-1)^2} = (x-\phi) (x+(\phi -1))$$
and if fact any algebraic set of 4 consecutive numbers just translates the graph left or right.
The way I recognized the factorization of the polynomial is by considering the coefficients of the 'golden polynomials.' The 'golden polynomials' are of the form:
$$(x^2 +x -1)^n$$
or
$$(x^2 -x -1)^n$$
After seeing this I questioned whether multiplying a different amount of consecutive numbers and adding 1 would still give 'golden polynomials' but unfortunately this was not true (however 2 consecutive numbers multiplied together then subtracted by 1 does give a 'golden polynomial')