Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l = 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum \(\pi\)(R + r)l sq.units
\(\frac { 22 }{ 7 } \times (4+1)\times 5\)
\(=\frac { 550 }{ 7 } \)
Therefore, C.S.A. = 78.57 cm²

Required Ratio = C.S.A. of the sphere: C.S.A. of the cylinder : C.S.A. of the cone
\(4\pi { r }^{ 2 }:2\pi rh:\pi rl,\ (l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 2r }^{ 2 } } =\sqrt { 2r } units)\)
\(=4:2:\sqrt { 2 } =2\sqrt { 2 } :\sqrt { 2 } :1\).

Let r be the radius of the hemisphere.
Given that, base area = \(\pi\)r² = 1386 sq. m
T.S.A. = 3 \(\pi\)r² sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m².

Let r₁ and r₂ be the radii of the balloons.
Given that, \(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 12 }{ 16 } =\frac { 3 }{ 4 } \)
Now, ratio of C.S.A. of balloons \(=\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } \)
Therefore, ratio of C.S.A. of balloons is 9:16.

Let r be the radius of the sphere. Given that, surface area of sphere = 154 m²
4\(\pi\)r² = 154
\(4\times \frac { 22 }{ 7 } \times { r }^{ 2 }=154\)
gives \({ r }^{ 2 }=154\times \frac { 1 }{ 4 } \times \frac { 7 }{ 22 } \)
hence, \({ r }^{ 2 }=\frac { 49 }{ 4 } \)We get r = \(\frac{7}{2}\)
Therefore, diameter is 7 m

Given that, radius r = 7 cm
Now, total surface area of the cone = \(\pi\)r(l + r)sq. units
T.S.A = 704 cm²
704 = \(\frac{22}{7}\times7(l+7)\)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Given that, height h = 25 Given that, height h = 25 m; thickness = 2 m.
internal radius r = 6 m
Now, external radius R = 6 + 2 = 8m
C.S.A. of the cylindrical tunnel = C.S.A. of the hollow cylinder
C.S.A. of the hollow cylinder = 2\(\pi\)(R + r)h sq.units 
\(2\times \frac { 22 }{ 7 } (8+6)\times 25\)
Hence, C.S.A. of the cylindrical tunnel = 2200 m²
Area covered by one litre of paint = 10 m²
Number of litres required to paint the tunnel \(=\frac { 2200 }{ 10 } =220\)
Therefore, 220 litres of paint is needed to paint the tunnel.

Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m²
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m²

Given that, C.S.A. of the cylinder = 88 sq. cm
2\(\pi\)rh = 88
\(2\times \frac { 22 }{ 7 } \times 14=88\) (given h = 14cm)
2r = \(\frac { 88\times 7 }{ 22\times 14 } =2\)
Therefore, diameter = 2 cm

Given that, height of the cylinder h = 20 cm ; radius r =14 cm
Now, C.S.A. of the cylinder = 2p\(\pi\)h sq. units
C.S.A. of the cylinder = \(2\times \frac { 22 }{ 7 } \times 14\times 20=2\times 22\times 2\times 20\)
T.S.A. of the cylinder \(=2\pi r(h+r)\)sq.units
\(=2\times \frac { 22 }{ 7 } \times 14\times (20+14)=2\times \frac { 22 }{ 7 } \times 14\times 34\)
= 2992 cm²
Therefore, C.S.A. = 1760 cm² and T.S.A. = 2992 cm²