First we use that the sum of $k$ Poisson($\lambda$) has distribution Poisson($k \lambda$). With this, we get that $Y|T\sim$Poisson($T\lambda$) and therefore $\mathbb{E}[Y|T]=T \lambda$.
Now, let us compute the variance. As you mentioned, we can use the law of total variance
$$
Var[Y] = Var[\mathbb{E}[Y|T]] + \mathbb{E} [Var[Y|T]].
$$
Notice that $Var[\mathbb{E}[Y|T]] = Var[\lambda T]=\lambda^2 Var[T]= \lambda^2 \theta$.
Now, for the second part,
$$
\mathbb{E} \left[Var[Y|T]\right] =
\mathbb{E} \left[\mathbb{E}[Y^2|T]\right] - 2 \mathbb{E} \left[\mathbb{E}[Y|T]^2\right]+E[T^2].
$$
It should be clear how to compute the second and third terms by using the distribution of $T$ and $\mathbb{E}[Y|T]$.
For the first term, notice that
\begin{align*}
\mathbb{E}[Y^2|T=t] &= \sum_{i,j=1}^t \mathbb{E} [X_i X_j]
\\&= \sum_{1\le i\neq j\le t} \mathbb{E} [X_i]\mathbb {E}[ X_j]
+\sum_{i=1}^t \mathbb{E} [X_i^2]
\\&= \sum_{1\le i\neq j\le t} \mathbb{E} [X_i]^2
+\sum_{i=1}^t \mathbb{E} [X_i^2]
\\&= t(t-1)(\lambda^2)
+t (\lambda^2+\lambda),
\end{align*}
where we used that $X_i's$ are i.i.d in the second and third identity. Therefore, we have $\mathbb{E}[Y^2|T]= T(T-1)\lambda^2 + T(\lambda^2+\lambda)$. Plugin everything back should give you the result. Let me know if you need further help.