a) To show that the total surface area is \( S = \frac{82}{x} + 4\pi x^2 \):
The solid consists of a cylinder with radius \( x \) cm and height \( h \) cm, and two hemispheres at each end.
Cylinder lateral surface area:
\( 2 \times \pi \times x \times h \)

Surface area of two hemispheres (each has surface area \( 2\pi x^2 \)):
\( 2 \times 2\pi x^2 = 4\pi x^2 \)

Total surface area:
\( S = 2 \times \pi \times x \times h + 4\pi x^2 \)

Given cylinder volume:
\( \pi \times x^2 \times h = 41 \)
\( h = \frac{41}{\pi x^2} \)

Substitute \( h \):
\( S = 2 \times \pi \times x \times \frac{41}{\pi x^2} + 4\pi x^2 \)
\( = \frac{2 \times 41}{x} + 4\pi x^2 \)
\( = \frac{82}{x} + 4\pi x^2 \)

Thus:
\( S = \frac{82}{x} + 4\pi x^2 \) [3]

b) (i) To find an expression for \( \frac{dS}{dx} \):
\( S = \frac{82}{x} + 4\pi x^2 = 82 x^{-1} + 4\pi x^2 \)
Differentiate:
\( \frac{dS}{dx} = 82 \times (-1) \times x^{-2} + 4\pi \times 2x \)
\( = -\frac{82}{x^2} + 8\pi x \)

(ii) To find the exact value of \( a \):
Set \( \frac{dS}{dx} = 0 \):
\( -\frac{82}{x^2} + 8\pi x = 0 \)
\( 8\pi x = \frac{82}{x^2} \)
\( 8\pi x^3 = 82 \)
\( x^3 = \frac{82}{8\pi} = \frac{41}{4\pi} \)
\( x = \left( \frac{41}{4\pi} \right)^{1/3} \)

Thus:
\( a = \left( \frac{41}{4\pi} \right)^{1/3} \) [5]

c) (i) To find an expression for \( \frac{d^2S}{dx^2} \):
\( \frac{dS}{dx} = -\frac{82}{x^2} + 8\pi x \)
Differentiate:
\( \frac{d^2S}{dx^2} = \frac{d}{dx} \left( -82 x^{-2} + 8\pi x \right) \)
\( = -82 \times (-2) x^{-3} + 8\pi \)
\( = \frac{164}{x^3} + 8\pi \)

(ii) To justify that \( S \) is a minimum when \( x = a \):
Evaluate \( \frac{d^2S}{dx^2} \) at \( x = a \), where \( a^3 = \frac{41}{4\pi} \):
\( \frac{d^2S}{dx^2} = \frac{164}{\left( \frac{41}{4\pi} \right)} + 8\pi \)
\( = \frac{164 \times 4\pi}{41} + 8\pi \)
\( = \frac{656\pi}{41} + 8\pi \approx 16\pi + 8\pi = 24\pi \)
Since \( 24\pi > 0 \), \( S \) has a local minimum at \( x = a \).

(iii) To find the minimum surface area:
Substitute \( x = a = \left( \frac{41}{4\pi} \right)^{1/3} \) into \( S \):
\( S = \frac{82}{\left( \frac{41}{4\pi} \right)^{1/3}} + 4\pi \left( \frac{41}{4\pi} \right)^{2/3} \)
\( = 82 \left( \frac{4\pi}{41} \right)^{1/3} + 4\pi \left( \frac{41}{4\pi} \right)^{2/3} \)
Numerically:
\( a \approx \left( \frac{41}{4 \times 3.1416} \right)^{1/3} \approx 1.483 \)
\( S \approx \frac{82}{1.483} + 4 \times 3.1416 \times 1.483^2 \)
\( \approx 55.292 + 27.622 \approx 82.914 \approx 82.9 \)

Thus:
Minimum surface area \( \approx 82.9 \) cm\(^2\) (to 3 significant figures) [6]