= Solution
Write the lower block of $T(a)$ as $B=\left(\begin{smallmatrix}0&b\\c&0\end{smallmatrix}\right)$. Then
$$
B^2=bcI_2,\qquad T(a)^2=\operatorname{diag}(1,bc,bc),\qquad T(a)^4=\operatorname{diag}(1,(bc)^2,(bc)^2).
$$
A representation of $C_4$ must satisfy its defining relation $T(a)^4=I_3$, so $(bc)^2=1$. For real $b,c$, this means $bc=1$ or $bc=-1$; it also ensures invertibility.
If $bc=-1$, then $T(a)^2\ne I_3$ while $T(a)^4=I_3$. The image of the generator has order four, so its four powers are distinct and the representation is faithful. If $bc=1$, then $T(a)^2=I_3$ but $T(a)\ne I_3$, since the lower block has zero diagonal. The image has order two and the kernel is $\{I,a^2\}$. Hence
$$
\boxed{\text{faithful exactly when }bc=-1;\qquad\text{unfaithful representation exactly when }bc=1.}
$$
For other values there is \b[no representation of $C_4$], rather than an additional unfaithful case; in particular $bc=0$ would make the assigned matrix singular. These conclusions are the <faithfulness criterion for a block representation of the cyclic group of order four>. There is no requirement that the proposed representation be orthogonal, so no separate conditions $|b|=|c|=1$ should be imposed.
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