Hint: First, start off with the constrained maximization problem and then use the Lagrangian method to get the first-order conditions. Then use the Kuhn-Tucker method (see Mathematics for Economist by Simon and Blume) to solve for your optimal demands.
The consumer solves the following maximization problem:
\begin{align*}
\max_{q_1,q_2} \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) \\
\text{s.t. } p_1 q_1 + p_2q_2 \le m.
\end{align*}
The Lagrangian associated with maximization problem is
\begin{equation*}
\mathcal{L} = \alpha_1q_1 + \alpha_2q_2 - \frac{1}{2}(q_1^2+q_2^2+2\epsilon q_1 q_2) + \lambda[m - p_1 q_1 - p_2q_2]
\end{equation*}
Then take the derivatives of $\mathcal{L}$ wrt to $q_{1}$, $q_{2}$, and $\lambda$ to get the following first-order conditions:
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial \lambda}=m-p_{1}q_{1}-p_{2}q_{2}\geq0
\end{equation}
Then we use the Kuhn-Tucker method:
$\lambda[m-p_{1}q_{1}-p_{2}q_{2}] = 0$ is the complementary slackness condition on $\lambda$
and
$\lambda\geq0$ is the non-negativity constraint on $\lambda$.
We can have two cases: either $\lambda>0$ or $\lambda=0$. I show what happens for $\lambda>0$ and then you can check what happens for $\lambda=0$.
If $\lambda>0$ then $m-p_{1}q_{1}-p_{2}q_{2}=0$ by the complementary slackness condition. In other words, this is the case where the budget constraint is binding, that is, $m=p_{1}q_{1}-p_{2}q_{2}$. Then we check whether the first-order conditions and the non-negativity condition hold.
After doing so we are left with the following set of equations:
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{1}}=\alpha_{1}-q_{1}+\epsilon q_{2} -\lambda p_{1} =0
\end{equation}
\begin{equation}
\frac{\partial \mathcal{L}}{\partial q_{2}}=\alpha_{2}-q_{2}+\epsilon q_{1} -\lambda p_{2}=0
\end{equation}
\begin{equation}
m-p_{1}q_{1}-m_{2}q_{2}=0
\end{equation}
Then substitute away $\lambda$ from the first two equations and solve for $q_{1}$ and $q_{2}$ using the binding budget constraint.
Edit: For the unconstrained problem you have no budget constraint so your set up is conceptually wrong. The way you get an unconstrained problem from a constrained problem is that you substitute away your budget constraint in the utility function. Also I haven't imposed non-negative conditions on $q_{1}$ and $q_{2}$, as then the process would be quite tedious.
$$
L(x,\lambda) = x_1^{\frac 34}x_2^{\frac 14}+\lambda_1(w x_3-17 x_1)+\lambda_2(24-x_1-x_2-x_3)
$$
$$
\nabla L =
\left\{\begin{array}{rcl}
\frac{3 \sqrt[4]{\text{x2}}}{4 \sqrt[4]{\text{x1}}}-17 \lambda_1-\lambda_2&=&0 \\
\frac{\text{x1}^{3/4}}{4 \text{x2}^{3/4}}-\lambda_2&=&0 \\
\lambda_1 w-\lambda_2&=&0 \\
w x_3-17 x_1&=&0 \\
24-x_1-x_2-x_3&=&0 \\
\end{array}\right.
$$
with the solution
$$
\left[
\begin{array}{cccccc}
x_1 & x_2 & x_3 & \lambda_1 & \lambda_2 & U\\
\frac{18 w}{w+17} & 6 & \frac{306}{w+17} & -\frac{3^{3/4}}{4 \sqrt[4]{w} \sqrt[4]{(w+17)^3}} & -\frac{3^{3/4} w^{3/4}}{4
\sqrt[4]{(w+17)^3}} & 6\ 3^{3/4} \left(\frac{w}{w+17}\right)^{3/4}
\end{array}
\right]
$$
Best Answer
So far so good. Next you make the derivative of $\mathcal L(c,l,\lambda)$ w.r.t $\lambda$, which is just the constraint.
$$\frac{\partial\mathcal L(c,l,\lambda)}{\partial \lambda}=24w-pc-wl=0$$
In combination with $l^*= 24-\left(\frac{w}{p}\right)^{\eta}$ we obtain
$$24w-pc-w\cdot \left(24-\left(\frac{w}{p}\right)^{\eta} \right)=0$$
It remains to solve the equation for $c$. I don´t think that you need the next steps. So I hide them and you can compare them with your own steps at the end.