For integer $ n,\ a_n$ is difined by $ a_n\equal{}\int_0^{\frac{\pi}{4}} (\cos x)^ndx$.

(1) Find $ a_{\minus{}2},\ a_{\minus{}1}$.

(2) Find the relation of $ a_n$ and $ a_{n\minus{}2}$.

(3) Prove that $ a_{2n}\equal{}b_n\plus{}\pi c_n$ for some rational number $ b_n,\ c_n$, then find $ c_n$ for $ n<0$.
### Part (1): Finding \( a_{-2} \) and \( a_{-1} \)

1. **Calculate \( a_{-2} \):**
   \[
   a_{-2} = \int_0^{\frac{\pi}{4}} (\cos x)^{-2} \, dx = \int_0^{\frac{\pi}{4}} \sec^2 x \, dx
   \]
   We know that the integral of \(\sec^2 x\) is \(\tan x\):
   \[
   a_{-2} = \left[ \tan x \right]_0^{\frac{\pi}{4}} = \tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1
   \]

2. **Calculate \( a_{-1} \):**
   \[
   a_{-1} = \int_0^{\frac{\pi}{4}} (\cos x)^{-1} \, dx = \int_0^{\frac{\pi}{4}} \sec x \, dx
   \]
   The integral of \(\sec x\) is \(\ln |\sec x + \tan x|\):
   \[
   a_{-1} = \left[ \ln |\sec x + \tan x| \right]_0^{\frac{\pi}{4}} = \ln (\sec \frac{\pi}{4} + \tan \frac{\pi}{4}) - \ln (\sec 0 + \tan 0)
   \]
   Since \(\sec \frac{\pi}{4} = \sqrt{2}\) and \(\tan \frac{\pi}{4} = 1\):
   \[
   a_{-1} = \ln (\sqrt{2} + 1) - \ln 1 = \ln (\sqrt{2} + 1)
   \]

### Part (2): Finding the relation between \( a_n \) and \( a_{n-2} \)

1. **Express \( a_{n-2} \) in terms of \( a_n \):**
   \[
   a_{n-2} = \int_0^{\frac{\pi}{4}} (\cos x)^{n-2} \, dx
   \]
   Using integration by parts, let \( u = \tan x (\cos x)^n \) and \( dv = dx \):
   \[
   du = n (\cos x)^{n-1} \sin x \, dx \quad \text{and} \quad v = x
   \]
   Applying integration by parts:
   \[
   a_{n-2} = \left[ \tan x (\cos x)^n \right]_0^{\frac{\pi}{4}} + n \int_0^{\frac{\pi}{4}} (\cos x)^{n-1} \sin x \tan x \, dx
   \]
   Evaluating the boundary term:
   \[
   \left[ \tan x (\cos x)^n \right]_0^{\frac{\pi}{4}} = \tan \frac{\pi}{4} (\cos \frac{\pi}{4})^n - \tan 0 (\cos 0)^n = 1 \cdot \left(\frac{1}{\sqrt{2}}\right)^n - 0 = \left(\frac{1}{\sqrt{2}}\right)^n
   \]
   Simplifying the integral:
   \[
   n \int_0^{\frac{\pi}{4}} (\cos x)^{n-1} \sin x \tan x \, dx = n \int_0^{\frac{\pi}{4}} (\cos x)^{n-2} (1 - \cos^2 x) \, dx
   \]
   \[
   = n \int_0^{\frac{\pi}{4}} (\cos x)^{n-2} \, dx - n \int_0^{\frac{\pi}{4}} (\cos x)^n \, dx
   \]
   \[
   = n a_{n-2} - n a_n
   \]
   Combining all terms:
   \[
   a_{n-2} = \left(\frac{1}{\sqrt{2}}\right)^n + n a_{n-2} - n a_n
   \]
   Solving for \( a_{n-2} \):
   \[
   a_{n-2} (1 - n) = \left(\frac{1}{\sqrt{2}}\right)^n - n a_n
   \]
   \[
   a_{n-2} = \frac{\left(\frac{1}{\sqrt{2}}\right)^n - n a_n}{1 - n}
   \]

### Part (3): Proving \( a_{2n} = b_n + \pi c_n \) and finding \( c_n \) for \( n < 0 \)

1. **Express \( a_{2n} \) in the form \( b_n + \pi c_n \):**
   \[
   a_{2n} = \int_0^{\frac{\pi}{4}} (\cos x)^{2n} \, dx
   \]
   Using the reduction formula:
   \[
   a_{2n} = \frac{2n}{2n-1} a_{2n-2} - \frac{1}{2n-1} \left(\frac{1}{\sqrt{2}}\right)^{2n}
   \]
   For \( n < 0 \), the term \(\left(\frac{1}{\sqrt{2}}\right)^{2n}\) becomes significant. We need to find the rational part \( c_n \):
   \[
   c_n = \frac{1}{2n-1}
   \]

The final answer is \( \boxed{ a_{-2} = 1 } \) and \( a_{-1} = \ln (\sqrt{2} + 1) \). The relation is \( a_{n-2} = \frac{\left(\frac{1}{\sqrt{2}}\right)^n - n a_n}{1 - n} \). For \( n < 0 \), \( c_n = \frac{1}{2n-1} \).