Answer: (a) \(\operatorname{sec}^2(x+4)-3\cos x\). (b) Approximately \(-0.138h\).

(a)

\(\frac{d}{dx}\tan(x+4)=\operatorname{sec}^2(x+4),\)

and

\(\frac{d}{dx}(-3\sin x)=-3\cos x.\)

Hence

\(\boxed{\frac{dy}{dx}=\operatorname{sec}^2(x+4)-3\cos x}.\)

(b) Write

\(y=\frac{\ln(2x+5)}{2e^{3x}}.\)

Using the quotient rule, with numerator \(u=\ln(2x+5)\) and denominator \(v=2e^{3x}\),

\(u'=\frac{2}{2x+5},\qquad v'=6e^{3x}.\)

So

\(\frac{dy}{dx} =\frac{2e^{3x}\cdot\frac{2}{2x+5}-6e^{3x}\ln(2x+5)}{4e^{6x}}.\)

This may be written as

\(\frac{dy}{dx}=\frac{\frac{2}{2x+5}-3\ln(2x+5)}{2e^{3x}}.\)

For a small increase \(h\),

\(\Delta y\approx h\frac{dy}{dx}.\)

At \(x=1\),

\(\frac{dy}{dx}\approx -0.138.\)

Therefore the approximate change in \(y\) is

\(\boxed{-0.138h}.\)