(1) $\mathrm{i}^{4 \mathrm{n}+3}$
(2) $i^{8 n+85}$
(3) $i^{16 n-14}$

Step 1 ：The problem is asking for the value of \(i^{4n+3}\), where \(i\) is the imaginary unit and \(n\) is an integer.

Step 2 ：The imaginary unit \(i\) is defined as \(\sqrt{-1}\), and it has a cyclic property: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). After \(i^4\), the cycle repeats.

Step 3 ：Therefore, we can simplify \(i^{4n+3}\) by using the cyclic property of \(i\).

Step 4 ：The function first calculates the exponent \(4n + 3\), then it calculates the cyclic exponent by taking the modulus of the exponent and 4.

Step 5 ：The function then returns the value of \(i\) to the power of the cyclic exponent.

Step 6 ：\(i^{4n+3}\) returns \(i\) if \(4n+3\) is congruent to 1 modulo 4, -1 if \(4n+3\) is congruent to 2 modulo 4, -\(i\) if \(4n+3\) is congruent to 3 modulo 4, and 1 if \(4n+3\) is congruent to 0 modulo 4.

Step 7 ：\(\boxed{\text{The value of } i^{4n+3} \text{ is given by the function, which returns } i \text{ if } 4n+3 \text{ is congruent to 1 modulo 4, -1 if } 4n+3 \text{ is congruent to 2 modulo 4, -} i \text{ if } 4n+3 \text{ is congruent to 3 modulo 4, and 1 if } 4n+3 \text{ is congruent to 0 modulo 4.}}\)