a) \( \sqrt{7^{2}}:(4+3)+21: 7-\sqrt{\sqrt[3]{27} \cdot 2}=8 \)

Let's break it down! First, we simplify each part of the equation: \( \sqrt{7^{2}} = 7 \). Next, we evaluate \( (4 + 3) = 7 \). Hence, \( 7 : 7 = 1 \). For the next part, \( 21 : 7 = 3 \). Lastly, we simplify \( \sqrt{\sqrt[3]{27} \cdot 2} \). The cube root of 27 is 3, so we have \( \sqrt{3 \cdot 2} = \sqrt{6} \). Now, putting it all together: \[ 1 + 3 - \sqrt{6} \] To conclude your original statement as follows, for it to be equal to 8, we can cleverly deduce that \( 1 + 3 = 4 \), and subtracting a small value like \( \sqrt{6} \) keeps it much lesser than 8. So, we investigate this as fundamentally incorrect! To double-check: If \( 1 + 3 - \sqrt{6} = 8 \), that is only valid if negating \( \sqrt{6} \) adds to a higher value, a paradoxical logic since \( \sqrt{6} < 4 \) leading us to incorrect arithmetic logic previously referenced, hence, concluding the statement is false because \( 4 - 2.45 \neq 8 \). This exercise gives us a glimpse into the beauty of arithmetic and logical deductions! Also, remember, it's always beneficial to double-check each step in a mathematical operation!