A demand function relates the quantity demanded of a good by a consumer with the price of the good. Thus we wish to find $Y = f(P_Y)$.
Setting up the optimization problem:
$$\max{U(X,Y)}$$
subject to: $$ I = P_x X + P_Y Y $$
where $I$ is income, $P_X$ is the price of good $X$, and $P_Y$ is the price of good $Y$.
Using the values you provided gives the optimization problem as:
$$ \max{ (XY + 10Y) } $$
subject to: $$ 100 = 1 \cdot X + P_Y Y $$
Setting this up as a Lagrange problem,
$$ L = XY + 10Y + \lambda (100 - X - P_Y Y )$$
Taking the first order conditions, we get:
$[X]:$ $\frac{ \partial U(X,Y) }{ \partial X} = Y - \lambda = 0$
$[Y]:$ $\frac{ \partial U(X,Y) }{ \partial Y} = X + 10 - \lambda P_Y = 0$
$[ \lambda ]:$ $\frac{ \partial U(X,Y) }{ \partial \lambda } = 100 - X - P_Y Y = 0$
Note, at this point you will usually take the second order conditions to ensure you have a maximum. Clearly you do have a maximum in this case since $U$ is strictly increasing in $X$ and $Y$.
Combining $[X]$ and $[Y]$ we get $X + 10 = Y P_Y$
We wish to get the demand for clothing, so we will solve for $X$ with the intention of substitution it into the budget constraint, $X = Y P_Y - 10$. Substituting into the constraint yields: $100 = 2 P_Y Y - 10$, or a final demand equation of:
$$ Y = \frac{45}{P_Y} $$
Finally, for a utility function to be quasi-linear, you must be able to express one utility as a linear function of one of the goods. Note in your case this may not be accomplished since you have an interaction between $X$ and $Y$. The reason quasi-linearity is nice is because it allows the expression of utility in terms of a numeraire good.
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.
Best Answer
In this case, $q_0$ is all of the "outside" goods, that is, the more it is consumed, the higher the utility will be. This is because the utility function is monotonically increasing in $q_0$, which suggests that the budget constraint should be \begin{equation} q_0 + p_1 q_1 + p_2 q_2 \le Y \end{equation} for budget level of $Y$, and prices of $p_1$ and $p_2$ for the good 1 and good 2. By using Lagrangian method or considering that a rational customer would spend all its budget, we can set $q_0 = Y- p_1 q_1 - p_2 q_2$. Plugging this back into the utility function, one needs to solve the following maximization problem \begin{equation} \max_{q_1,q_2} (Y- p_1 q_1 - p_2 q_2) + 5q_1+5q_2−\frac{1}{2}(q_1^2+q_2^2+2dq_1q_2). \end{equation} First order conditions then suggest that \begin{align} q_1^*(p_1,p_2) &= \frac{5(1-d)-p_1+dp_2}{1-d^2} \\ q_2^*(p_1,p_2) &= \frac{5(1-d)-p_2+dp_1}{1-d^2}. \end{align} These are well-known linear demand functions and the cross-price effects are given as \begin{align} \frac{\partial q_1}{ \partial p_2} = \frac{\partial q_2}{\partial p_1}=\frac{d}{1-d^2}. \end{align} Hence, these two goods are substitutes if $\frac{d}{1-d^2}>0$. They are complements if $\frac{d}{1-d^2}<0$.