[THOUGHT] Translate conditions into algebra. From (A), f(0)=f(2)=0, so f(x)=ax(x-2)*(x-r) with a>0. Continuity of g at t implies f(t)=0 so t is one of {0,2,r}. If t=r then g/(x(x-2)) becomes nonnegative a|x-r| so no negative limits, contradicting (B). If t=0 then for natural m we have g(1)=f(1)>0 making -(7/2)g(1)<0, impossible since it must be a natural number. Hence t=2. With t=2, for natural m the limit equals -a(m-r) for m<2 and a(m-r) for m>=2. To have exactly two negative m, r must be in (3,4], giving negative at m=2,3 only. Then {g(-1), -(7/2)g(1)} must equal {2,3}. Solve to find r and a, then compute g(-5). [/THOUGHT] [PYTHON] import sympy as sp

a, r = sp.symbols('a r', positive=True, real=True)

t = 2 case

g_minus1 = 3a(1+r) # derived g1 = a*(1-r) # since x=1<2 so g=-f, with f(1)=a(r-1) expr2 = -sp.Rational(7,2)*g1 # -(7/2)g(1) = (7/2)a(r-1)

Solve the only feasible assignment: g(-1)=3, expr2=2

sol = sp.solve([sp.Eq(g_minus1, 3), sp.Eq(expr2, 2)], [a, r], dict=True) print(sol)

sol = sol[0] a_val = sp.simplify(sol[a]) r_val = sp.simplify(sol[r]) print("a =", a_val, "r =", r_val)

Compute g(-5): for -5<2, g(-5) = -f(-5) = 35a(5+r)

g_minus5 = sp.simplify(35a_val(5+r_val)) print("g(-5) =", g_minus5)

Verify the set condition on natural m: compute sign for m=1..6

def L(m): m=sp.Rational(m,1) if m < 2: return sp.simplify(-a_val*(m-r_val)) else: return sp.simplify(a_val*(m-r_val)) for m in range(1,7): val=L(m) print(m, val, "negative?" , sp.sign(val)<0)

print("g(-1) =", sp.simplify(g_minus1.subs(sol))) print("-(7/2)g(1) =", sp.simplify(expr2.subs(sol))) [/PYTHON][THOUGHT] Translate conditions into algebra. From (A), (f(0)=f(2)=0), so (f(x)=a,x(x-2)(x-r)) with (a>0). Continuity at (t) forces (f(t)=0), hence (t\in{0,2,r}).
If (t=r), then (\frac{g(x)}{x(x-2)}) becomes nonnegative everywhere, contradicting (B).
If (t=0), then (g(1)=f(1)>0) so (-\frac72 g(1)<0), impossible since it must be a natural number in (B). Thus (t=2).

With (t=2):
(g(-1)=3a(1+r)), (g(1)=a(1-r)), so (-\frac72 g(1)=\frac72 a(r-1)).
The negative-limit naturals are ({2,3}), so ({g(-1),-\frac72 g(1)}={3,2}). Solving gives (r=\frac{11}{3}), (a=\frac{3}{14}).
Then for (-5<2), (g(-5)=-f(-5)=35a(5+r)=65). (\boxed{65})

[THOUGHT] The Python output gives (g(-5)=65). Therefore the requested value is (\boxed{65}).