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Substituting $b=6$ into $ab-6a+5b=15$ we get the absurdity $30=15$.   Thus $b$ can never assume the value 6.   By algebraic manipulation we can express $a$ in terms of $b$ as $a = (15-5b)/(b-6)$.    Now, $15-5b > 0$ iff $b < 3$ and $b-6 > 0$ iff $b > 6$.   Therefore, $a$ is positive only when $3 < b < 6$.   Among these two plausible values of $b$ (4 and 5), only $b=5$ turns $a$ into a positive integer $a=10$.  Thus, $|a-b|$ has only one value 5 and it must be the minimum.

It seems there's not enough information to find $f(0)+f(4)$;  I need to know the value of $f(x)$ at one more point.

It turns out that the second constraint alone completely determines the values of $f(-1)$, $f(0)$, and $f(1)$.   The first constraint just makes it possible to determine the values of $f$ at other domain points.

We set $x = 0, 1, -1$ in order in the second constraint to get these systems of equations
\begin{eqnarray*}
    f(-1) &=& f(0)^2-2 \\
    f(0) &=& (f(1)-1)^2+1-2 \\
    f(0) &=& (f(-1)+1)^2+1-2
\end{eqnarray*}
Solving we find 3 solutions $(f(-1), f(0), f(1)) = (-1,-1,1)$ or $(-2,0,0)$ or $(-2,0,2)$.
The answer you are looking for is any one of $(-1,1)$, $(0,0)$, or $(0,2)$.

It turns out that the second constraint alone completely determines the values of $f(-1)$, $f(0)$, and $f(1)$.   The first constraint just makes it possible to determine the values of $f$ at other domain points.

We set $x = 0, 1, -1$ in order in the second constraint to get these systems of equations
\begin{eqnarray*}
    f(-1) &=& f(0)^2-2 \\
    f(0) &=& (f(1)-1)^2+1-2 \\
    f(0) &=& (f(-1)+1)^2+1-2
\end{eqnarray*}
Solving we find 3 solutions $(f(-1), f(0), f(1)) = (-1,-1,1)$ or $(-2,0,0)$ or $(-2,0,2)$.
The answer you are looking for is any one of $(-1,1)$, $(0,0)$, or $(0,2)$.

I think the question contains a typo: $f^{-1}(x)$ should have the form $(ax+b)/(cx+d)$ instead.

Since $70 = 2\cdot 5\cdot 7$, there are 8 ways to factor 70 into two factors (not taking order into consideration): $1\cdot 70$, $2\cdot 35$, $5\cdot 14$, $7\cdot 10$, and four other ways similar to these but where both factors are negative.  By checking all possibilities under the constraints $3b+c=41$ and $bc=70$, we find that only $b=2$ and $c=35$ satisfy both constraints.    Substituting $b=2$ into the first constraint $2a+3b=52$ gives $a=23$.  So $a+b+c = 23+2+35 = 60$.

Perhaps you meant to say $a^2+b^2+c^2+ab+ac+bc=81$ instead.  If that's the case, then there are three solutions:
    (3, 4, 4)
    (4, 3, 4)
    (4, 4, 3)
 

Property 1 & 2 can be combined into
\[
    xf(x) + yf(y) = 1+(x+y)f(x+y)
\]
for all $x\in S$ and $y\in S$ such that $x+y\in S$.  Taking $x=y=1$, the combined property asserts that $2f(1) = 1+2f(2)$.  Taking $x=-1$, Property 1 asserts that $f(-1) = -f(-1)$; thus $f(-1) = 0$.  Taking $x=2$ and $y=-1$, the combined property asserts that $2f(2)=1+f(1)$.  Solving the system of two equations
\begin{eqnarray*}
    2f(1) &=& 1+2f(2) \\
    2f(2) &=& 1+f(1)
\end{eqnarray*}
gives $f(1) = 2$ (with $f(2)=3/2$ but we don't need to know that).  Thus $n\times s = 1\times 2 = 2$.

Let $q(x)$ be the quotient polynomial and let $r(x)$ be the remainder polynomial when $f(x)$ is divided by $(x-1)(x-2)$.  We have
\[
    f(x) = q(x)(x-1)(x-2) + r(x).
\]
Since $(x-1)(x-2)$ is quadratic, we can write $r(x)$ as $ax+b$ for some constants $a$, $b$.   From $f(1)=2$ and $f(2)=8$ we get
\begin{eqnarray*}
    2 = f(1) = r(1) = a+b \\
    8 = f(2) = r(2) = 2a+b
\end{eqnarray*}
Solving, we get $a = 6$ and $b = -4$, so the remainder $r(x)=6x-4$.

Note that $\sqrt{x^2}$ is equivalent to $|x|$ so the question asks how many real $x$ makes $\sqrt{1200-|x|}$ an integer.  Each such $x$ satisfies $1200 - |x| \ge 0$, and $1200-|x|$ must be a square number.  Since $34^2 = 1156$ and $35^2  = 1225$, the possible values of $1200-|x|$ are $0^2, 1^2, 2^2, \ldots, 34^2$.  Each of these square numbers gives exactly 2 values for $x$, one positive and one negative.   E.g.
\[
0=1200-|1200|=1200-|-1200|
\]
and
\[
34^2=1200-|44|=1200-|-44|.
\]
So the answer is $35*2=70$.