Answer: \(\mathrm f(x)=\frac14(3x+5)^{4/3}+4x+12\).

Set up the integral using the correct limits or antiderivative first. When definite integration is used, evaluate the upper limit and lower limit separately before subtracting.

Integrate \(\mathrm f''(x)=(3x+5)^{-2/3}\):

\(\mathrm f'(x)=(3x+5)^{1/3}+C\).

Using \(\mathrm f'(1)=6\),

\(6=8^{1/3}+C=2+C\), so \(C=4\).

Thus

\(\mathrm f'(x)=(3x+5)^{1/3}+4\).

Integrate again:

\(\mathrm f(x)=\frac14(3x+5)^{4/3}+4x+D\).

Using \(\mathrm f(1)=20\),

\(20=\frac14(8)^{4/3}+4+D\).

Since \(8^{4/3}=16\), this gives

\(20=4+4+D\), so \(D=12\).

Therefore

\(\mathrm f(x)=\frac14(3x+5)^{4/3}+4x+12\).

This gives the required answer: \(\mathrm f(x)=\frac14(3x+5)^{4/3}+4x+12\).

Answer: \(y=\frac1{10}(3x+2)^{\frac53}+2x-1\).

Integrate the second derivative to find the gradient:

\(\frac{dy}{dx}=\int (3x+2)^{-\frac13}\,dx.\)

Using the reverse chain rule,

\(\frac{dy}{dx}=\frac12(3x+2)^{\frac23}+C.\)

The gradient is \(4\) when \(x=2\), so

\(4=\frac12(8)^{\frac23}+C.\)

Since \(8^{\frac23}=4\),

\(4=2+C,\)

and therefore \(C=2\). Thus

\(\frac{dy}{dx}=\frac12(3x+2)^{\frac23}+2.\)

Integrate again:

So

\(y=\frac1{10}(3x+2)^{\frac53}+2x+D.\)

The curve passes through \((2,6.2)\), so

\(6.2=\frac1{10}(8)^{\frac53}+4+D.\)

Since \(8^{\frac53}=32\),

\(6.2=3.2+4+D,\)

which gives \(D=-1\). Therefore

\(y=\frac1{10}(3x+2)^{\frac53}+2x-1.\)

Answer: \(y=\frac13(2x+3)^{3/2}+2x-4\); normal \(x+5y-58=0\).

Integrate the second derivative:

\(\frac{dy}{dx}=\int(2x+3)^{-1/2}\,dx.\)

Since \(\frac{d}{dx}(2x+3)^{1/2}=(2x+3)^{-1/2}\),

\(\frac{dy}{dx}=(2x+3)^{1/2}+C.\)

When \(x=3\), the gradient is \(5\). Therefore

\(5=(9)^{1/2}+C=3+C,\)

so \(C=2\). Hence

\(\frac{dy}{dx}=(2x+3)^{1/2}+2.\)

Integrate again:

\(y=\frac13(2x+3)^{3/2}+2x+D.\)

Use the point \(\left(\frac12,-\frac13\right)\). When \(x=\frac12\), \(2x+3=4\), so

\(-\frac13=\frac13(4)^{3/2}+1+D.\)

Thus

\(-\frac13=\frac83+1+D=\frac{11}{3}+D,\)

so \(D=-4\). The curve is

\(\boxed{y=\frac13(2x+3)^{3/2}+2x-4}.\)

At \(x=3\),

\(y=\frac13(9)^{3/2}+6-4=9+2=11.\)

The gradient of the tangent is \(5\), so the gradient of the normal is \(-\frac15\). Therefore

\(y-11=-\frac15(x-3).\)

Multiplying by \(5\) gives

\(5y-55=-x+3,\)

so the normal is

\(\boxed{x+5y-58=0}.\)

Answer: (i) \(\dfrac5{14}(7x-10)^{\frac25}+C\); (ii) \(a=\dfrac{253}{7}\).

For part (i), increase the power by \(1\):

\(-\dfrac35+1=\dfrac25.\)

Also divide by the derivative of \(7x-10\), which is \(7\). Therefore

\(\int(7x-10)^{-\frac35}\,dx=\dfrac{(7x-10)^{\frac25}}{7\cdot\frac25}+C.\)

So

\(\int(7x-10)^{-\frac35}\,dx=\dfrac5{14}(7x-10)^{\frac25}+C.\)

For part (ii), use the result from part (i):

\(\dfrac5{14}\left[(7x-10)^{\frac25}\right]_6^a=\dfrac{25}{14}.\)

So

\(\dfrac5{14}(7a-10)^{\frac25}-\dfrac5{14}(7\cdot6-10)^{\frac25}=\dfrac{25}{14}.\)

Since \(7\cdot6-10=32\) and \(32^{\frac25}=4\),

\(\dfrac5{14}(7a-10)^{\frac25}-\dfrac{20}{14}=\dfrac{25}{14}.\)

Hence

\(\dfrac5{14}(7a-10)^{\frac25}=\dfrac{45}{14}.\)

So

\((7a-10)^{\frac25}=9.\)

Raise both sides to the power \(\dfrac52\):

\(7a-10=9^{\frac52}=243.\)

Therefore

\(7a=253,\)

so

\(\boxed{a=\dfrac{253}{7}}.\)

Answer: (i) \(-\dfrac{15}{(3x+2)^2}\), (ii) \(-\dfrac{10}{3x+2}+C\), (iii) \(\dfrac34\).

Write

\(\dfrac5{3x+2}=5(3x+2)^{-1}.\)

Then

\(\dfrac{d}{dx}\left(5(3x+2)^{-1}\right)=5(-1)(3x+2)^{-2}\cdot3.\)

So

\(\dfrac{d}{dx}\left(\dfrac5{3x+2}\right)=-\dfrac{15}{(3x+2)^2}.\)

Since

\(\dfrac{d}{dx}\left(\dfrac5{3x+2}\right)=-\dfrac{15}{(3x+2)^2},\)

multiplying by \(-2\) gives

\(\dfrac{d}{dx}\left(-\dfrac{10}{3x+2}\right)=\dfrac{30}{(3x+2)^2}.\)

Therefore

\(\int\dfrac{30}{(3x+2)^2}\,dx=-\dfrac{10}{3x+2}+C.\)

For the definite integral,

\(\int_1^2\dfrac{30}{(3x+2)^2}\,dx=\left[-\dfrac{10}{3x+2}\right]_1^2.\)

So

\(-\dfrac{10}{8}-\left(-\dfrac{10}{5}\right)=-\dfrac54+2=\dfrac34.\)

Hence

\(\boxed{\dfrac34}.\)