There seems to be something amiss about this question.
Assuming fn to mean the nth derivative with respect to x, where n=0⇒f(x) and integer n≥0:
fn(x)=-36sin(6x)=-62sin(6x)
fn+1(x)=-63cos(6x)
fn+2(x)=64sin(6x)=-62fn(x)
fn+3(x)=65cos(6x)=-62fn+1(x).
Therefore we can write, in general: fn+2(x)=-62fn(x). This means that it doesn't matter whether n is even or odd.
If we let y=fn(x), then y"=-62y, that is, y"+62y=0 which has the general solution:
y=Asin(6x)+Bcos(6x), implying fn(x)=Asin(6x)+Bcos(6x) (A and B are constants).
We can also write fn(x)=-62fn-2(x). This is an iterative DE, so sooner or later fn-2(x) is the same as f0(x)=f(x).
[NOTE: If n=2, f"(x)=-36f(x) and f(x)=Asin(6x)+Bcos(6x).]
fn(0)=-62fn-2(0). If we consider the initial conditions we have 2 consecutive derivatives (n=0 and n=1) and in each case we have a non-zero number.
When we consider fn and fn+1 (two consecutive derivatives), we can see that the variable term is sin(6x) or cos(6x). Now, when x=0 we get either fn(0)=0 and fn+1(0)≠0 or we get fn(0)≠0 and fn+1(0)=0. So it's not possible to have f(0)≠0 and f'(0)≠0. We have also seen the iterative DE as the general solution, which involves both sin(6x) and cos(6x).
If f(x)=Asin(6x)+Bcos(6x), f(0)=B=-5; f'(0)=6A=6, so A=1 and f(x)=sin(6x)-5cos(6x).
If this reasoning is correct f(π/4)=-1, because sin(6x)=sin(3π/2)=-1 and cos(3π/2)=0.
This satisfies the initial conditions and, when n=2, also satisfies:
f2(x)=-36f(x); f"(x)=-36sin(6x)-180cos(6x).