= Solution
The <Weierstrass M-test> gives a continuous periodic <Weierstrass function>, since $\sum a^{-k}<\infty$. For $5^m\le n<5^{m+1}$, the given best approximant leaves the tail
$$
g-t_n=\sum_{k=m+1}^\infty a^{-k}\cos(5^kx).
$$
Its <supremum norm> is at most the sum of its coefficients, and at $x=0$ all cosine terms equal one, attaining that bound. Consequently
$$
\boxed{E_n(g)=\sum_{k=m+1}^\infty a^{-k}=\frac{a^{-m}}{a-1}.}
$$
Put $\alpha=\ln a/\ln5$. Then $a^{-m}=(5^m)^{-\alpha}$ and $n/5^m\in[1,5)$, giving $E_n(g)\asymp n^{-\alpha}$, with constants depending only on $a$.
For $1<a<5$, part (a) gives $\omega(g,1/n)\le C_an^{-\alpha}$. The <first Jackson theorem for periodic approximation>, $E_n(g)\le C\omega(g,1/n)$, supplies the matching lower bound. Hence the <modulus of continuity> has the precise order
$$
\boxed{\omega(g,\delta)\asymp\delta^{\ln a/\ln5},\qquad1<a<5,\quad\delta\downarrow0.}
$$
At $a=5$, the <inverse theorem for trigonometric approximation> gives the upper bound $C\delta\ln(1/\delta)$, and part (c) proves the matching lower bound. Thus the <critical Weierstrass modulus> is
$$
\boxed{\omega(g,\delta)\asymp\delta\ln(1/\delta),\qquad a=5.}
$$
To pass from scales $1/n$ to all small $\delta$, use monotonicity of the <modulus of continuity> and adjacent reciprocals, which are comparable to $\delta$.
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