The first question to ask is when is $f(x)^{g(x)} = 1$
Well we know that $a^0 = 1$ so $g(x)=0$ gives up to 2 real solutions since $g(x)$ is a quadratic. (*)
Next we know that $1^a = 1$ so $f(x)=0$ gives up to 2 real solutions since $f(x)$ is a quadratic (**)
Now the final (most difficult insight to spot) is $(-1)^{2a}=1$ (where $a$ is an integer) so $f(x)=-1$ and $g(x)$ is even gives up to 2 solutions since they're quadratics (***)
Now time to look at the main question:
Solve the equation
$$(x^2 -5x +5)^{(x^2 -11x +30)}=1$$
(*) Tells us that $x^2 -11x + 30 = (x-6)(x-5)=0$ so $x=6$ and $x=5$ are solutions
(**) Tells us that $x^2-5x+5=1$ so $x^2 -5x +4 = (x-4)(x-1)=0$ so $x=4$ and $x=1$ are solutions.
(***) Tells us that $x^2 -5x +5=-1$ so $x^2 -5x +6 = (x-3)(x-2)=0$ so $x=3$ and $x=2$ are (potential) solutions. Now lets check that $g(x)=x^2 -11x +30$ is even in both cases. $g(3) = odd + odd + even=even$ and $g(2)=even + even + even = even$.
Remark: Interestingly enough for some quadratic $q(x)$ of the from $q(x) = (2a+1)x^2 + (2b+1)x + 2c$ (where $a,b,c$ are integers) $q(x)$ is even for all $x$ in the integers as explained before. We can use this fact later. (****)
Question 1:
Solve the equation
$$(x^2 -7x +11)^{(x^2 -13x +42)}=1$$
We can use the same method we were using before to see that
$x^2 - 13x +42 =(x-6)(x-7)=0$ so $x=6$ and $x=7$ are solutions
$x^2-7x+11=1$ so $x^2 -7x +10 = (x-2)(x-5)=0$ so $x=2$ and $x=5$ are solutions
$x^2 - 7x +11=-1$ so $x^2-7x+12 = (x-3)(x-4) = 0$ so $x=3$ and $x=4$ are (potential) solutions. Using (****) the exponent $g(x)$ must be even.
Question 2:
We are trying to write a problem with solutions
$$n,(n+1),(n+2),(n+3),(n+4),(n+5)$$
Looking from the 2 questions above we can notice that (*) generates the solutions $(n+4),(n+5)$, then (**) generates the solutions $n,(n+3)$ and finally (***) generates the solutions $(n+1),(n+2)$.
In fact we should explain why the these statements always work.
By using factor theorem, the first statement should give $g(x)$ in the form (****) and $g(x) = (x-(n+4))(x-(n+5)) = x^2 -(2n+9)x +(n+4)(n+5)$ which is of the correct from (we can quite clearly see that $(n+4)(n+5)$ is even as its $even \times odd$).
For the second and third statement we have that $f(x)-1=(x-n)(x-(n+3))= x^2 -(2n+3)x +(n^2 + 3n)$ and $f(x)+1=(x-(n+1))(x-(n+2))= x^2 -(2n+3)x +(n^2 + 3n +2)$ which both rearrange to the same $f(x)=x^2 -(2n+3)x +(n^2 + 3n+1)$
Now returning to the actual focus of the question we know that $g(x) = x^2 - (2n+9)x +(n^2+9n+20)$ and $f(x) = x^2 -(2n+3)x + (n^2 +3n +1)$
To get the set of solutions $3,4,5,6,7,8$ we can set $n=3$ to give the equation
$$(x^2 -9x + 19)^{(x^2 -15x +56)}=1$$
To get the set of solutions $4,5,6,7,8,9$ we can set $n=4$ to give the equation
$$(x^2 -11x + 29)^{(x^2 -17x +72)}=1$$
Question 3:
For the equation
$$(x^2 -5x +5)^{(x+2)(x-2)}=1$$
Using (*) we get the solutions $x=-2$ and $x=2$
Using (**) we get the solutions $x=4$ and $x=1$
Using (***) we get that $x=2$ but we have already used this solution and then we get $x=3$ but this then makes exponent $g(x)$ odd so this means that $x=3$ isn't a solution.
Question 4:
For the equation
$$(x^2-6x+10)^{(x+2)(x-1)}=1$$
Using (*) we get the solutions $x=-2$ and $x=1$
Using (**) we get a repeated root of $x=3$
Using (***) we find that we need to solve $x^2 -6x +11=0$ but this has a discriminant of $36-44=-8$ which gives non-real solutions.
Question 5:
Its actually quite simple to find an equation with only 2 solutions. We can do this in multiple ways but I'll name a few. Where we are solving $f(x)^{g(x)}=1$ where they're quadratics then we can give both $f(x)=1$ and $g(x)=0$ repeated roots (and then $f(x)=-1$ has no real roots) for example
$$(x^2-6x+10)^{(x^2-4x+4)}=1$$
With solutions $x=2,3$
We could also give $g(x)=0$ two distinct roots and both $f(x)=1$ and $f(x)=-1$ no real roots for example
$$(x^2-6x+11)^{(x^2+x-2)}=1$$
With solutions $x=-2,1$
Now to find an equation with 5 solutions I first thought we could give $g(x)=0$ repeated roots but then $g(x)$ is no longer in the form (****) so this isn't viable. However what we could do is give $f(x)=-1$ repeated roots and then make $g(x)$ in the form in (****) for example
$$(x^2-6x+8)^{(x^2+x-2)}=1$$
(Even though this won't give all integer solutions)
We could also make it so 1 of the solutions achieved from either case (*) or (**) or (***) be repeated for example
$$(x^2 -7x +11)^{(x^2 -11x +30)}=1$$
With solutions $x=2,3,4,5,6$