1. Evaluate \( \sec^2 \left( \tan^{-1} \frac{1}{2} \right) \):
Let \( \theta = \tan^{-1} \frac{1}{2} \), which implies \( \tan \theta = \frac{1}{2} \). Construct a right triangle where the opposite side is 1 and the adjacent side is 2. The hypotenuse is calculated as \( \sqrt{1^2 + 2^2} = \sqrt{5} \).
Then, \( \sec^2 \theta = \frac{\text{Hypotenuse}^2}{\text{Adjacent}^2} = \frac{5}{4} \).
2. Evaluate \( \csc^2 \left( \cot^{-1} \frac{1}{3} \right) \):
Let \( \phi = \cot^{-1} \frac{1}{3} \), which implies \( \cot \phi = \frac{1}{3} \). Construct a right triangle where the adjacent side is 1 and the opposite side is 3. The hypotenuse is calculated as \( \sqrt{1^2 + 3^2} = \sqrt{10} \).
Then, \( \csc^2 \phi = \frac{\text{Hypotenuse}^2}{\text{Opposite}^2} = \frac{10}{9} \).
3. Add the results:
\( \sec^2 \left( \tan^{-1} \frac{1}{2} \right) + \csc^2 \left( \cot^{-1} \frac{1}{3} \right) = \frac{5}{4} + \frac{10}{9} \).
To add these fractions, find a common denominator:
\( \frac{5}{4} + \frac{10}{9} = \frac{5 \times 9}{4 \times 9} + \frac{10 \times 4}{9 \times 4} = \frac{45}{36} + \frac{40}{36} = \frac{45 + 40}{36} = \frac{85}{36} \).
Final Answer: \( \boxed{\frac{85}{36}} \)