Solution: 1. Let $f,g\in V$ and $a\in \mathbb{R} $, then we need to show that $T(af+g) = aT(f)+T(g)$. Observe that \begin{align*} T(af+g) & = \frac{\partial (af+g)}{\partial x} \\ & = \frac{\partial af}{\partial x} + \frac{\partial g}{\partial x}\\ & = a\frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}\\ & = a T(f) + T(g). \end{align*}

2. We need to find the null space of $T$. Recall that \begin{align*} \text{Null}(T) & = \left\{ f\in V: T(f) = 0 \right\} \\ & = \left\{ f\in V: \frac{\partial f(x,y)}{\partial x} = 0 \right\} \\ & = \left\{ f\in V: f(x,y) = F(y) \right\} \\ & = \text{all polynomials in variable $y$ only}. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\text{Null space of $T$ = set of all polynomials in variable $y$ }}} \]

3. We need to find the range of $T$. Recall that \begin{align*} \text{Range}(T) & = \left\{ T(f):f\in V \right\} . \end{align*} We claim that the range of $T$ is $V$. For this, take $f\in V$. Let \[ f(x,y) = \sum_{i=0,j=0}^{i=m,j=n} a_ib_jx^i y^j \] Define \[ g(x,y) = \int f(x,y) ~\mathrm{d}x. \]

Note that \begin{align*} \int f(x,y)~\mathrm{d} x & = \int \left( \sum_{i=0,j=0}^{i=m,j=n} a_ib_jx^i y^j \right) \mathrm{d} x\\ & = \sum_{i=0,j=0}^{i=m,j=n} \int a_ib_jx^i y^j~\mathrm{d} x\\ & = \sum_{i=0,j=0}^{i=m,j=n} \frac{a_ib_jx^{i+1} y^j}{i+1} \end{align*} Then it is clear that \begin{align*} T(g) & = \frac{\partial }{\partial x} \left( \int f(x,y)~\mathrm{d} x \right) \\ & = \frac{\partial }{\partial x} \left( \sum_{i=0,j=0}^{i=m,j=n} \frac{a_ib_jx^{i+1} y^j}{i+1} \right) \\ & = \sum_{i=0,j=0}^{i=m,j=n} \frac{a_ib_jx^{i} y^j}{i+1}\times (i+1) \\ & = \sum_{i=0,j=0}^{i=m,j=n} a_ib_jx^i y^j \\ & = f(x,y). \end{align*} Thus, the range of $T$ is $V$.