Let the diameter be x and area be y.
To find y when x = 82, use Newton's forward interpolation form
∴ x0 + nh = 82 ⇒ 80 + n(5) ⇒ 82 - 80 = 2
⇒ n = \(\frac{2}{5}\) = 0.4
The difference table is

∴ y(82) = y0 + \(\frac { n }{ 1! } { \triangle y }_{ 0 }+\frac { n(n+1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n+1)(n+2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }\)+ \(\frac { n(n+1)(n+2)(n-3) }{ 4! } ​​{ \triangle }^{ 4 }{ y }_{ 0 }​​​​​​​​​​​​​​\)

+ \(\frac{(0.4)(0.4-1)(0.4-2)}{6}\)(-2)

= 5026 + 259.2 + (0.4) (-0.6) (20) + \(\frac{(0.4)(-0.6)(-1.6)(-1)}{3}\) + \(\frac{(0.4)(-0.6)(-1.6)(-2.6)}{3}\)
= 5026 + 259.2 - 4.8 - 0.128 - 0.1664
= 5280.10
∴ When the diameter is 82, area of circle is 5280.1 (≅ 5281)
To find y when x = 91, use Newton's backward interpolation formula.
∴ x_n + nh = 91 ⇒ 100 + n(5) = 91
⇒ 5n = 91 - 100
⇒ 5n = -9 ⇒ n = \(\frac{-9}{5}\) = 1.8
Newton's backward interpolation formula is
y_(x=xn+nh) = yn + \(\frac { n }{ 1! } \triangledown { y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }​​​​​​​​​​​​​​\)
y(x = 91) = 7854 - 1.8 (766) +

\(+\frac{(-1.8)(-1.8+1)(-1.8+2)(-1.8+3)}{3!}\)(4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + (-1.8) (-0.8) (20) + (-1.8) (-0.8) (0.2) + \(\frac{(-1.8)(-0.8)(0.2)(1.2)}{6}\) (4)
⇒ y_{(x = 91)} = 7854 - 1378.8 + 28.8 + 0.288 + 0.2304
⇒ y_{(x = 91)} = 6504.5
Hence when the diameter is 91, area is 6504.5 ≅ 6504

Since only four values are given,
(E -1)4 u₀ = 0
⇒ (E⁴ - 4E³ + 6E² - 4E + 1) u₀
= u₄ - 4 u₃ + 6 u₂ - 4 u₁ + u₀
⇒ 385 - 4(u₃) + 6(520) - 4 (556) + 560 =0
⇒ 385 - 4u₃ + 3120 - 2224 + 560 = 0
⇒ 1841 - 4u₃ = 0
⇒ 1841 = 4u₃
⇒ u₃ = \(\frac{1841}{4}\) = 460.25
∴ u₃ = 460.25

Given

Here the intervals are unequal
∴ By Lagranges interpolation formula, we have
x₀ = 0, x₁ = 1, x₂ = 3, x₃ = 4
y₀ = -12, y₁ = 0, y₂ = 6, y₃ = 12 and x = x.
∴ y = f(x) = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)
= \(\frac { (x-1)(x-3)(x-4) }{ (0-1)(0-3)(0-4) } (-12)+\frac { (x-0)(x-3)(x-4) }{ (1-0)(1-3)(1-4) } (0)+\frac { (x-0)(x-1)(x-4) }{ (3-0)(3-1)(3-4) } (6)+\frac { (x-1)(x-3)(x-4) }{ (4-0)(4-1)(4-3) } (12)\)
= \(\frac { (x-1)(x-3)(x-4) }{ (-1)(-3)(-4) } (-12)+0+\frac { x(x-1)(x-4) }{ (3)(2)(-1) } (6)+\frac { x(x-1)(x-3) }{ (4)(3)(1) } (12)\)
= +[(x - 1)(x - 3)(x - 4)] - x (x - 1)(x -  4) + x(x - 1)(x - 3)
= +[(x³-4x+3)(x-4)] -x(x²-5x+4) + x(x²-4x + 3)
= - (x³ - 8x²+ 19x - 12) - 4x² + 3x
= (x - 4)(x² - 4x + 3) - x (x² - 5x + 4) + x(x² - 4x + 3)

= x³ - 7x² + 19x - 12.

Let the missing entries be y₂ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero
⇒ (E - 1)⁴y_k = 0
⇒ (E₄- 4E₃ + 6E₂ - 4E + 1) y_k = 0  (1)
Put k = 0 in (1) we get,
y₄ - 4y₃ + 6y₂ - 4y₁ + y₀ = 0
y₄ - 4(18) + 6y₂ - 4(11) + 7 = 0
⇒ y₄ - 72 + 6y₂ - 44 + 7 = 0
⇒ y₄ + 6y₂ = 109    (2)
Put k = 1 in (1) we get,
(E₄- 4E₃ + 6E₂ - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ +y₁ = 0
⇒ 32 - 4 (y₄) + 6 (18) - 4y₂ + 11 = 0
32 - 4y₄ + 108 - 4y₂ + 11 = 0
⇒ -4y₄ - 4y₂ + 151 = 0

Adding,           
⇒ y₂ = 14.25
Substituting y₂ = 14.25 in (2) we get,
y₄ + 6 (14.25) = 109
⇒ y₄ + 85.5 = 109
⇒ y₄ = 109 - 58.5
⇒ y₄ = 23.5

Given:

Here, the intervals are unequal:
∴ By Lagrange's interpolation formula, we have
x₀ = 1982, x₁ = 1983, x₂ = 1984, x₃ = 1986
y₀ = 150, y₁ = 235, y₂ = 365, y₃ = 525 and x = 1985
=\(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } \times { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })(x_{ 1 }-{ x }_{ 3 }) } \times { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } \times { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 2 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \times { y }_{ 3 }\)  \(\frac { (1985-1983)(1985-1984)(1985-1986) }{ (1982-1983)(1982-1984)(1982-1986) } \times 150+\frac { (1985-1982)(1985-1984)(1985-1986) }{ (1983-1982)(1983-1984)(1983-1986) } \times 235+\frac { (1985-1982)(1985-1983)(1985-1986) }{ (1984-1982)(1984-1983)(1984-1986) } \times 365+\frac { (198 5-1982)(1985-1983)(1985-1984) }{ (1986-1982)(1986-1983)(1986-1984) } \times 525\)

\(=\frac { (2)(1)(-1) }{ (-1)(-2)(-4) } \times 150+\frac { (3)(1)(-1) }{ (1)(-1)(-3) } \times \) 
= 37.5 + 235 + 547.5 + 131.25
= 481.25