I'll try to explain this.
As all the functions are continuous, all change of signs must occur only at the zeroes of the function,hence you mark all the zeroes in the real axis and you know that any change of sign can only occur at those zeroes due to continuity. Now Two conditions arise:
The zero is non-repeating. It means that $f'(x) \neq 0$ at that point. Hence the function must change sign. To clarify it,consider that function doesn't changes sign ,then X-axis is a tangent and hence slope of tangent is $0$.
So,you just plug one value (calculating sign of function at which is easier ) and find sign of function in that corresponding to that interval, then you just alternate the signs for all other intervals if the corresponding zeroes are non repeating.
The zero is repeating. I am not very sure what will happen then. So I'll suggest you to find the values in next interval manually.
Now if you know the sign of function in each interval, to draw approximate graph,draw a free hand sketch. Note that this method does not tell you anything about values in intervals except whether they are positive or negative, so there you are on your own.
I'll try the first one:
$-(x+5)(x+2)(1-x)(2x-8)=(x+5)(x+2)(x-1)(2x-8)$ with zeroes at $-5,-2,1,4$. Now, as $f(x) > 0$ for $x=100$ we just alternate signs of $f(x)$ at each zero. Hence $f(x) > 0$ in the intervals $(4, \infty ),(-2,1),(- \infty,-5)$ and $f(x) < 0$ for other intervals.
Note that there was nothing magical about $100$, you could use any number (though I think it would be better if you choose a number very loo or very high,i.e, corresponding to either $(- \infty ,a) $ or $(b,\infty )$. And then just alternate signs in intervals. Also you might want to watch out for repeated roots as I am not sure what to do. Just to be sure, check the next interval again.
In 5,check for value at say $-111$ it is positive. Now the function is positive in $(\infty, -7),(2,3)$ and negative elsewhere.