The value of $$\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5}\times\left(\frac{7}{8}-\frac{5}{4}\right)\right]$$ is:

$$\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5}\times\left(\frac{7}{8}-\frac{5}{4}\right)\right]=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5}\times\left(\frac{7-10}{8}\right)\right]$$

$$=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\frac{1}{5}\times\left(\frac{-3}{8}\right)\right]$$

$$=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}-\left(\frac{-3}{40}\right)\right]$$

$$=\frac{33}{40}+\frac{1}{5}\left[\frac{4}{5}+\frac{3}{40}\right]$$

$$=\frac{33}{40}+\frac{1}{5}\left[\frac{32+3}{40}\right]$$

$$=\frac{33}{40}+\frac{1}{5}\left[\frac{35}{40}\right]$$

$$=\frac{33}{40}+\frac{7}{40}$$

$$=\frac{40}{40}$$

$$=1$$

Hence, the correct answer is Option C