For point A, you just have to replace t by a in the given function, like this
[tex]\begin{gathered} G\mleft(t\mright)=\mleft(3t-5\mright)^2+4t-1 \\ \text{ Replacing} \\ G\mleft(a\mright)=\mleft(3a-5\mright)^2+4a-1 \\ \text{ Solving you have} \\ G(a)=(3a-5)(3a-5)+4a-1 \\ G(a)=9a^2-30a+25+4a-1 \\ \text{ Add similar terms} \\ G(a)=9a^2-26a+24 \end{gathered}[/tex]
For point B, you just have to replace t by a+2 in the given function, like this
[tex]\begin{gathered} G(t)=(3t-5)^2+4t-1 \\ \text{ Replacing} \\ G(a+2)=(3(a+2)-5)^2+4(a+2)-1 \\ \text{ Solving you have} \\ G(a+2)=(3a+6-5)^2+4(a+2)-1 \\ G(a+2)=(3a+1)^2+4a+8-1 \\ G(a+2)=(3a+1)(3a+1)+4a+8-1 \\ G(a+2)=9a^2+6a+1+4a+8-1 \\ \text{ Add similar terms} \\ G(a+2)=9a^2+10a+8 \end{gathered}[/tex]
