time scaling and shifting of cosine in Fourier transform
Using the shifting and scaling properties of the Fourier transform is a rather complicated way of computing the Fourier transform of $\cos(at+b)$. A more straightforward way is to realize that

$$\cos(at+b)=\frac12\left[e^{jat}e^{jb}+e^{-jat}e^{-jb}\right]\tag{1}$$

From the Fourier transform pair

$$e^{jat}\Longleftrightarrow 2\pi\delta(\omega-a)\tag{2}$$

we immediately obtain

$$\cos(at+b)\Longleftrightarrow \pi\left[e^{jb}\delta(\omega-a)+e^{-jb}\delta(\omega+a)\right]\tag{3}$$

But having said that, your result is correct. You just need to simplify it appropriately. In order to do so, you need to know two things. First,

$$\delta\left(\frac{\omega}{a}\right)=|a|\delta(\omega)\tag{4}$$

and, second,

$$f(\omega)\delta(\omega-\omega_0)=f(\omega_0)\delta(\omega-\omega_0)\tag{5}$$

for any $f(\omega)$ that is continuous at $\omega_0$.

From $(4)$ you get

$$\delta(\omega/a-1)=|a|\delta(\omega-a)\tag{6}$$

And from $(5)$ you get

$$e^{j\omega b/a}\delta(\omega-a)=e^{jb}\delta(\omega-a)\tag{7}$$

Applying similar transformations to the second term in your result leads to the form given in $(3)$.