Let $T:\mathbb C^2\rightarrow\mathbb C^2$ defined by $T(a,b)=(2a+\iota b,a+2b)$
Solution: Let $T^*$ be the ajoint of $T,$then
$\langle T(a,b),(c,d)\rangle =\langle(a,b),T^*(c,d)\rangle$
$\implies\langle (a,b),T^*(c,d)\rangle=\langle (2a+\iota b,a+2b),(c,d)\rangle$
$\implies\langle (a,b),T^*(c,d)\rangle=\bar c(2a+\iota b)+\bar d(a+2b)=2a\bar c+\iota b\bar c+a\bar d+2b\bar d=a(2\bar c+\bar d)+b(\bar c\iota+\bar d)$
$\implies \langle (a,b),T^*(c,d)\rangle=\langle (a,b),(2\bar c+\bar d,\bar c\iota+\bar d)\rangle$
$\implies T^*(c,d)=(2\bar c+\bar d,\bar c\iota+\bar d)$
To Check the normality of $T,$ we have to show that $TT^*=T^*T$
Now,$TT^*(a,b)=T(2\bar a+\bar b,\bar a\iota+2\bar b)=(2(2\bar a+\bar b)+\iota(\bar a\iota+2\bar b),(2\bar a+\bar b)+2(\bar a\iota+\bar b))=(3\bar a+2\bar b(1+\iota),2\bar a(1+\iota)+5\bar b)$
and
$T^*T(a,b)=T^*(2a+\iota b,a+2b)=(2(2\bar a-\iota \bar b)+(a+2b),(2\bar a+\iota \bar b)\iota+2(\bar a+2\bar b))=(5\bar a+2\bar b(1-\iota),2\bar a(1+\iota)+3\bar b)$
Hence,$$(3\bar a+2\bar b(1+\iota),2\bar a(1+\iota)+5\bar b)\neq(5\bar a+2\bar b(1-\iota),2\bar a(1+\iota)+3\bar b)$$
$$TT^*\neq T^*T$$
$T$ is not normal
But, in the answer key it is given that $T$ is normal
Please check my calculations
Your computation strategy is fine, but I see two minor mistakes in your derivation of $T^*$.
First, you just lost a 2 at one point. You should instead get $$ \implies\langle (a,b),T^*(c,d)\rangle=\bar c(2a+\iota b)+\bar d(a+2b)=2a\bar c+\iota b\bar c+a\bar d+2b\bar d=a(2\bar c+\bar d)+b(\bar c\iota+\color{red}2\bar d). $$
Second, a couple steps later you forgot to conjugate some stuff when pulling inside an inner product. Here's the fixed version of the step in question (also includes the extra 2 from the first mistake): $$\begin{align*} \cdots &\implies\langle (a,b),T^*(c,d)\rangle = a(2\bar c+\bar d)+b(i \bar c+\color{red}2\bar d) \\ &\implies \langle (a,b),T^*(c,d)\rangle=\langle (a,b),(2c+d,-ic+\color{red}2d)\rangle \end{align*}$$
If you make these fixes and follow the same method then you'll get the expected result. To help you check your computations, I'll say:
$$ \begin{align} T^*(c,d) &= (2c + d, ic + 2d)\\ TT^*(a,b) = T^*T(a,b) &= \Big(5a + (2+2i)b, (2-2i)a + 5b\Big). \end{align} $$
Bonus fact: If you write $T$ and $T^*$ as matrices, then $T^*$ will always be the conjugate transpose of $T$. Check here and ctrl+f for both "adjoint" and "conjugate transpose". If you're taking a class and you haven't learned this fact then you shouldn't rely on it in your official written solution, but you can still use it to sanity check your work. It's much easier for me to find out where your mistakes are when I just already know we should be getting $T^*(a,b) = (2a+b, ia+2b)$.