Solution:
Let the numbers be a and b
Then,
ab = 17
⇒ a = 1 and b = 17
So,
$$\eqalign{ = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr = \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} \cr = \frac{{{1^2} + {{\left( {17} \right)}^2}}}{{{{\left( {1 \times 17} \right)}^2}}} \cr = \frac{{290}}{{289}} \cr} $$

Solution:
Let the fraction be $$\frac{x}{y}$$
Then,
$$\eqalign{ \Leftrightarrow \frac{{x + 2}}{{y + 3}} = \frac{7}{9} \cr \Leftrightarrow 9x - 7y = 3.....(i) \cr {\text{And,}} \cr \Leftrightarrow \frac{{x - 1}}{{y - 1}} = \frac{4}{5} \cr \Leftrightarrow 5x - 4y = 1.....(ii) \cr} $$
Solving (i) and (ii), we get :
x = 5 and y = 6
Hence, the original fraction is $$\frac{5}{6}$$

Solution:
Let the number be x
Then,
$$\eqalign{ \Leftrightarrow x \times \frac{2}{3}x = 864 \cr \Leftrightarrow \frac{2}{3}{x^2} = 864 \cr \Leftrightarrow {x^2} = \left( {\frac{{864 \times 3}}{2}} \right) \cr \Leftrightarrow {x^2} = 1296 \cr \Leftrightarrow x = \sqrt {1296} \cr \Leftrightarrow x = 36 \cr} $$

Solution:
Let the denominator be x
Then, numerator = x + 5
Now,
$$\eqalign{ \Leftrightarrow \frac{{x + 5}}{x} - \frac{{x + 5}}{{x + 5}} = \frac{5}{4} \cr \Leftrightarrow \frac{{x + 5}}{x} = \frac{5}{4} + 1 \cr \Leftrightarrow \frac{{x + 5}}{x} = \frac{9}{4} \cr \Leftrightarrow \frac{{x + 5}}{x} = 2\frac{1}{4} \cr} $$
So, the fraction is $$2\frac{1}{4}$$

Solution:
Let the middle digit be x
Then,
2x = 10 or x = 5
So, the number is either 253 or 352
Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.
Hence, required number = 253

Solution:
Let the numbers be a, b and c.
Then, a² + b² + c² = 138
And (ab + bc + ca) = 131
∴ (a + b + c)²
= a² + b² + c² + 2(ab + bc + ca)
= 138 + 2 × 131
= 400
⇒ (a + b + c) = $$\sqrt {400} $$  = 20

Solution:
Let the number be x
Then,
$$\eqalign{ \Leftrightarrow \frac{x}{5} + 4 = \frac{x}{4} - 10 \cr \Leftrightarrow \frac{x}{4} - \frac{x}{5} = 14 \cr \Leftrightarrow \frac{x}{{20}} = 14 \cr \Leftrightarrow x = 14 \times 20 \cr \Leftrightarrow x = 280 \cr} $$