1. Empty
2. Solution: False. Consider $1=(z^m+1)+(-z^m)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$. Note that \[(z^m+1)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}\]and\[-z^m\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$ is not closed under addition. Hence it is not a subspace of $\ca P(\mb F)$.
3. Solution: False. Consider $z=(z^{2}+z)+(-z^2)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}$. Note that \[(z^2+z)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}\]and\[-z^2\in \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}$ is not closed under addition. Hence it is not a subspace of $\ca P(\mb F)$.
4. Solution: Define $p\in\ca P(\mb F)$ by \[p(z)=(z-\lambda_1)^{n-m+1}(z-\lambda_2)\cdots(z-\lambda_m).\]Then $p$ is a polynomial of degree $n$ such that $0=p(\lambda_1)=\cdots=p(\lambda_{m})$ and such that $p$ has no other zeros.
5. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 2.
6. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 4.
7. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 5.
8. Solution: First we show that $T$ is a linear map. Then we show $Tp \in\ca P(\R)$ for a basis of $p \in\ca P(\R)$, then by linearity of $T$, we have $Tp \in\ca P(\R)$ for every polynomial $p \in\ca P(\R)$. For any $\lambda\in\R$ and $p,q\in\ca P(\R)$, we have \begin{align*} T(p+q)=&\dfrac{(p+q)-(p+q)(3)}{x-3}=\dfrac{(p+q)-p(3)-q(3)}{x-3}\\ =&\dfrac{p-p(3)}{x-3}+\dfrac{q-q(3)}{x-3}=Tp+Tq, \end{align*} if $x\ne 3$. Similarly, \[T(\lambda p)=\dfrac{(\lambda p)-(\lambda p)(3)}{x-3}=\dfrac{\lambda p-\lambda p(3)}{x-3}=\lambda\dfrac{p-p(3)}{x-3}=\lambda Tp.\]If $x=3$, then $T$ is a composition of the differentiation map and evaluation map. Both of them are linear, hence $T$ is also linear. We can show it directly \[ T(\lambda p+q)=(\lambda p+q)'(3)=(\lambda p’+q’)(3)=\lambda p'(3)+q'(3)=\lambda Tp+Tq. \]Therefore $T$ is a linear map. Let us consider $Tx^n$ for $n\in \mb N^+$, if $x\ne 3$, \[T(x^n)=\frac{x^n-3^n}{x-3}=x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}\in\ca P(\R).\] Moreover, if $x=3$, we have $T(x^n)=3^{n-1}n$. Note that when $x=3$, it is true that\[ x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}=3^{n-1}n. \]We get \[T(x^n)=x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}\in\ca P(\R)\]for $x\in \R$. Similarly, we can show $T(1)=0\in \ca P(\R)$.
Since any polynomial of $\ca P(\R)$ is a linear combination of $1$, $x$, $x^2$, $\cdots$, it follows that $Tp \in\ca P(\R)$ for every polynomial $p \in\ca P(\R)$.
I am not sure the textbook indicates that $1$, $x$, $x^2$, $\cdots$ is a basis of $\ca P(\R)$, so I use some easier arguments such as any polynomial of $\ca P(\R)$ is a linear combination of $1$, $x$, $x^2$, $\cdots$.
9. Solution: If $f(z)=a_nz^n+\cdots+a_1z+a_0$, where $a_n,\cdots,a_0\in\C$, then \[\overline{f(\bar{z})}=\overline{a_n}z^n+\cdots+\overline{a_1}z+\overline{a_0}.\]That implies $\overline{f(\bar{z})}$ is a polynomial. As the product of polynomials is a polynomial as well, we conclude $q$ is a polynomial.
Now let us show $q$ has only real coefficients. Denote $q(z)$ by \[q(z)=\mu_{2n}z^{2n}+\cdots+\mu_1z+\mu_0.\]Note that $\overline{q(\bar{z})}=\overline{f(\bar{z})\overline{f(z)}}=\overline{f(\bar{z})}f(z)=q(z)$, it follows \[ \overline{\mu_{2n}}z^{2n}+\cdots+\overline{\mu_1}z+\overline{\mu_0}=\mu_{2n}z^{2n}+\cdots+\mu_1z+\mu_0. \]Hence $\overline{\mu_k}=\mu_k$, i.e $\mu_k\in \R$, for $k=0,\cdots,2n$.
Here you can also compute the coefficients $\mu_k$ in terms of $a_i$ and show $\mu_k=\overline{\mu_k}$ or use some calculus methods.
10. Note that $x_0,x_1,\cdots,x_m$ are distinct, we can define the polynomial\[f(x)=\sum_{j=0}^m\frac{(x-x_0)(x-x_1)\cdots(x-x_{j-1})(x-x_{j+1})\cdots(x-x_m)}{(x_j-x_0)(x_j-x_1)\cdots(x_j-x_{j-1})(x_j-x_{j+1})\cdots(x_j-x_m)}p(x_j).\]Then it is obvious that $f(x)\in\ca P_m(\C)$. Moreover, since $x_j$ and $p(x_j)$, $j=0,1,\cdots,m$, are real, it follows that the coefficients of $f(x)$ are real. Hence it suffices to show that $p(x)=f(x)$.
By plugging $x=x_i$ into $f(x)$, we have $f(x_i)=p(x_i)$ since all summands except one are zero (see the link in the remark below),
\[
\frac{(x_i-x_0)(x-x_1)\cdots(x_i-x_{j-1})(x_i-x_{j+1})\cdots(x_i-x_m)}{(x_j-x_0)(x_j-x_1)\cdots(x_j-x_{j-1})(x_j-x_{j+1})\cdots(x_j-x_m)}=\delta_{ij},
\]where $\delta_{ij}=0$ if $i\ne j$ and $\delta_{ij}=1$ if $i=j$.
This implies that $f-p$ has $m+1$ distinct zeros. Since $f-p\in\ca P_m(\C)$, it follows from 4.12 that the degree of $f-p$ can not be nonnegative. Hence $f-p$ is the zero polynomial, thus completing the proof.
The polynomial used here is called the Lagrange Interpolating Polynomial . Please see the following link for more detail.
11. By the division algorithm of polynomials in 4.8, we know that for every polynomial $f\in\ca P(\mb F)$ there exist unique polynomials $q$ and $r$ such that \[f=pq+r,\quad\text{and}\quad \deg r<\deg p.\]This implies that $\ca P(\mb F)=U\oplus \ca P_{\deg p-1}(\mb F)$.
Therefore\[\ca P(\mb F)/U\cong\ca P_{\deg p-1}(\mb F).\]It follows that $$\dim\ca P(\mb F)/U=\dim\ca P_{\deg p-1}(\mb F)=\deg p.$$Moreover, a basis of $\ca P(\mb F)/U$ is $1+U,x+U,x^2+U,\cdots,x^{\deg p -1}+U$.
Here I used the fact that if $V=U\oplus W$, then $V/U\cong W$. Please try to prove as the following alternative solution.
Another solution (explains the solution above more explicitly).
For any given polynomial $f\in\ca P(\mb F)$, let $r(f)$ be the reminder of $f$ divided by $q$. Note that $\deg r(f)<\deg q$, we have a map $r:\ca P(\mb F)\to \ca P_{\deg p-1}(\mb F)$. One can check this is a linear map. Moreover, $\mathrm{Null}\, r=U$. By taking polynomials in $\ca P_{\deg p-1}(\mb F)$, we have that $\mathrm{range}\,r=\ca P_{\deg p-1}(\mb F)$.
By 3.91(d), we have that \[\ca P(\mb F)/U=\ca P(\mb F)/\mathrm{Null}\, r\cong \mathrm{range}\,r=\ca P_{\deg p-1}(\mb F).\]Now the problem is solved similarly.
