Answer: (a) approximate change \(=13.1h\). (b) \(\dfrac{7-\sqrt2}{6}\).

(a) Let

\(y=\mathrm{e}^{3x+2}\tan x.\)

Using the product rule,

\(\frac{\mathrm{d}y}{\mathrm{d}x}=3\mathrm{e}^{3x+2}\tan x+\mathrm{e}^{3x+2}\sec^2x.\)

For a small increase \(h\) in \(x\),

\(\delta y\approx h\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0.1}.\)

Substituting \(x=0.1\),

\(\left.\frac{\mathrm{d}y}{\mathrm{d}x}\right|_{x=0.1}=13.0768\ldots.\)

Therefore the approximate change in \(y\) is

\(13.1h.\)

(b) Since

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\sin(3x+\pi),\)

integrate to get

\(y=-\frac13\cos(3x+\pi)+C.\)

The curve passes through \(\left(\frac{\pi}{9},\frac43\right)\), so

\(\frac43=-\frac13\cos\left(\frac{\pi}{3}+\pi\right)+C.\)

Since \(\cos\frac{4\pi}{3}=-\frac12\),

\(\frac43=\frac16+C,\)

so

\(C=\frac76.\)

When \(x=\frac{5\pi}{12}\),

\(3x+\pi=\frac{5\pi}{4}+\pi=\frac{9\pi}{4}.\)

Thus

\(y=-\frac13\cos\frac{9\pi}{4}+\frac76=-\frac{\sqrt2}{6}+\frac76=\frac{7-\sqrt2}{6}.\)