Recall some identities:
[tex]\cos^2x+\sin^2x=1[/tex]
[tex]\tan^2x+1=\sec^2x=\dfrac1{\cos^2x}[/tex]
[tex]\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta[/tex]
[tex]\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta[/tex]
Not sure what part (c) is asking for, but I assume it's [tex]\tan(\alpha+\beta)[/tex], in which case
[tex]\tan(\alpha+\beta)=\dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}[/tex]
If [tex]\tan\alpha=-\dfrac43[/tex], then
[tex]\dfrac1{\cos^2\alpha}=1+\left(-\dfrac43\right)^2=\dfrac{25}9[/tex]
[tex]\implies\cos\alpha=\pm\dfrac35[/tex]
We know that [tex]\alpha[/tex] lies in quadrant 2, i.e. [tex]\dfrac\pi2<\alpha<\pi[/tex], so we expect [tex]\cos\alpha<0[/tex]. So we take the negative root. We also find that
[tex]\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\iff-\dfrac43=\dfrac{\sin\alpha}{-\frac35}\implies\sin\alpha=\dfrac45[/tex]
If [tex]\cos\beta=\dfrac23[/tex], then
[tex]\sin^2\beta=1-\left(\dfrac23\right)^2=\dfrac59\implies\sin\beta=\pm\dfrac{\sqrt5}3[/tex]
Since [tex]\beta[/tex] lies in quadrant 1, i.e. [tex]0<\beta<\dfrac\pi2[/tex], we know that [tex]\sin\beta>0[/tex], so we take the positive root.
Now,
[tex]\cos(\alpha+\beta)=-\dfrac35\cdot\dfrac23-\dfrac45\cdot\dfrac{\sqrt5}3=-\dfrac52-\dfrac4{3\sqrt5}[/tex]
[tex]\sin(\alpha+\beta)=\dfrac45\cdot\dfrac23+\left(-\dfrac35\right)\cdot\dfrac{\sqrt5}3=\dfrac8{15}-\dfrac1{\sqrt5}[/tex]
Then it follows that
[tex]\tan(\alpha+\beta)=\dfrac{\frac8{15}-\frac1{\sqrt5}}{-\frac52-\frac4{3\sqrt5}}=\dfrac{54-25\sqrt5}{22}[/tex]
