I guess you mean [tex]\cos a=-\dfrac25[/tex]. Since [tex]a[/tex] is in quadrant III, we expect [tex]\sin a<0[/tex]. Then
[tex]\sin a=-\sqrt{1-\cos^2a}=-\dfrac{\sqrt{21}}5[/tex]
Since [tex]b[/tex] is in quadrant I, we expect [tex]\sin b>0[/tex], so that
[tex]\sin b=\sqrt{1-\cos^2b}=\dfrac{\sqrt{15}}4[/tex]
Now,
[tex]\sin(a+b)=\sin a\cos b+\cos a\sin b=-\dfrac{2\sqrt{15}+\sqrt{21}}{20}[/tex]
[tex]\cos(a+b)=\cos a\cos b-\sin a\sin b=\dfrac{3\sqrt{35}-2}{20}[/tex]
and
[tex]\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}=-\dfrac{2\sqrt{15}+\sqrt{21}}{3\sqrt{35}-2}[/tex]
