Finding the fourier series coefficients of the signal $$1+\sin (\omega_0 t) + \cos (\omega_0 t) + \cos (2\omega_0 t + \pi / 4) $$
Because as you rightly point out it is already given in something very close to a Fourier series, there is no need to perform the whole procedure. Just rearrange to get it in exactly the right format.

I think you've over-cooked it with the summation pattern. There are just three frequency terms (DC, \$\omega_0\$ and \$2\omega_0\$) so you can just list them individually.

You can rearrange your second equation thusly:

$$
1 + cos(\omega_{0}t) + sin(\omega_{0}t) + \frac{1}{\sqrt{2}}cos(2\omega_{0}t) - \frac{1}{\sqrt{2}}sin(2\omega_{0}t)
$$

And then read the coefficients straight off:

$$
a_0 = 1, a_1 = 1, a_2 = \frac{1}{\sqrt{2}}, b_1 = 1, b_2 = -\frac{1}{\sqrt{2}}
$$

Note some texts prefer \$f(x) = \frac{1}{2}a_0 + ...\$ as their Fourier series, in which case \$a_0 = 2\$.