I'm trying to solve the following question (the answer is $var(T) = N$).
I'm getting a different result
$Var(T) = E(T^2) - [E(T)]^2]$
$T = X - Y$
$\implies Var(T) = E(X^2-2XY + Y^2)-[E(X)-E(Y)]^2 = E(X^2)-2E(XY) + E(Y^2) - [E(X)]^2 + 2E(X)E(Y) - [E(Y)]^2$
Now set $Y = N -X$.
Therefore
$E(XY) = N E(X) - E(X^2)$
$E(Y) = N - E(X)$
$E(Y^2) = N^2 - 2N*E(X) + E(X^2)$
Therefore,
$Var(T) = E(X^2) - 2[N*E(X) - E(X^2)] + N^2 - 2NE(X) + E(X^2) - [E(X)]^2 + 2E(X)[N-E(X)] - [N-E(X)]^2 = 4E(X^2) - 4[E(X)]^2 = 4*VAR(X)$
Am I wrong? Why does $Var(T) = N$? Thanks.

You are almost there. Observe that $X$ is a binomial random variable with $p=1/2$ (I presume). Therefore, $var(X)=Np(1-p)=N/4$ and $var(T)=4var(X)=N$.
Note that the calculations could be simplified significantly by observing that $$T = 2X-N \implies var(T)=4var(X)-0=N.$$