We know that:
[tex]\cos (\alpha-\beta)=\sin (\alpha)\sin (\beta)+\cos (\alpha)\cos (\beta)[/tex]Let's take a closer look at alpha. We know that:
[tex]\begin{gathered} \tan ^2(\alpha)=\sec ^2(\alpha)-1 \\ \rightarrow\tan ^2(\alpha)=\frac{1}{\cos^2(\alpha)}-1 \\ \\ \rightarrow\tan ^2(\alpha)+1=\frac{1}{\cos^2(\alpha)} \\ \\ \rightarrow\cos ^2(\alpha)=\frac{1}{\tan^2(\alpha)+1} \\ \\ \rightarrow\cos ^{}(\alpha)=\pm\text{ }\sqrt[]{\frac{1}{\tan^2(\alpha)+1}} \end{gathered}[/tex]We also know that alpha is in quadrant II, where cosine is negative. This way,
[tex]\begin{gathered} \cos ^{}(\alpha)=-\text{ }\sqrt[]{\frac{1}{\tan^2(\alpha)+1}} \\ \\ \rightarrow\cos ^{}(\alpha)=-\text{ }\sqrt[]{\frac{1}{(-\frac{12}{5})^2+1}} \\ \\ \Rightarrow\cos (\alpha)=-\frac{5}{13} \end{gathered}[/tex]We also know that:
[tex]\tan (\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)}\rightarrow\sin (\alpha)=\tan (\alpha)\cos (\alpha)[/tex]This way,
[tex]\begin{gathered} \sin (\alpha)=(-\frac{12}{5})(-\frac{5}{13}) \\ \\ \Rightarrow\sin (\alpha)=\frac{12}{13} \end{gathered}[/tex]We can conclude that:
[tex]\begin{gathered} \sin (\alpha)=\frac{12}{13} \\ \cos (\alpha)=-\frac{5}{13} \end{gathered}[/tex]Now, let's take a closer look at beta. We know that:
[tex]\begin{gathered} \sin ^2(\beta)=1-\cos ^2(\beta) \\ \rightarrow\sin (\beta)=\pm\text{ }\sqrt[]{1-\cos^2(\beta)} \end{gathered}[/tex]And since beta lies in quadrant 4, where sine is negative,
[tex]\begin{gathered} \sin (\beta)=-\text{ }\sqrt[]{1-\cos^2(\beta)}\rightarrow\sin (\beta)=-\text{ }\sqrt[]{1-(\frac{3}{5})^2} \\ \\ \Rightarrow\sin (\beta)=-\frac{4}{5}_{} \end{gathered}[/tex]This way, we can conclude that
[tex]\begin{gathered} \sin (\beta)=-\frac{4}{5} \\ \cos (\beta)=\frac{3}{5} \end{gathered}[/tex]Now, we'll use the identity from the begining:
[tex]\cos (\alpha-\beta)=\sin (\alpha)\sin (\beta)+\cos (\alpha)\cos (\beta)[/tex]With the values we've calculated:
[tex]\begin{gathered} \sin (\alpha)=\frac{12}{13} \\ \cos (\alpha)=-\frac{5}{13} \\ \text{and} \\ \sin (\beta)=-\frac{4}{5} \\ \cos (\beta)=\frac{3}{5} \end{gathered}[/tex]This way,
[tex]\begin{gathered} \cos (\alpha-\beta)=\sin (\alpha)\sin (\beta)+\cos (\alpha)\cos (\beta) \\ \\ \rightarrow\cos (\alpha-\beta)=(\frac{12}{13})(-\frac{4}{5})+(-\frac{5}{13})(\frac{3}{5}) \\ \\ \Rightarrow\cos (\alpha-\beta)=-\frac{63}{65} \end{gathered}[/tex]Therefore, we can conclude that:
[tex]\cos (\alpha-\beta)=-\frac{63}{65}[/tex]