Since $T:\mathbb{R}^3\to\mathbb{R}^2$, then you are looking for a 2 by 3 matrix. If you don't know already,
$e_1=\begin{bmatrix}1\\0\\0\end{bmatrix},\qquad e_2=\begin{bmatrix}0\\1\\0\end{bmatrix},\qquad e_3=\begin{bmatrix}0\\0\\1\end{bmatrix}$
Since $T(e_1)=\begin{bmatrix}3\\1\end{bmatrix}$, and $T(e_1)=T\begin{bmatrix}1\\0\\0\end{bmatrix}=$ the first column of $T$ (try multiplying out to see why), then you know that the first column of $T$ is $\begin{bmatrix}3\\1\end{bmatrix}$
Then repeat this reasoning for the other columns to get $T$.
Once you know what $T$ is, then use the definition of matrix-vector multiplication to answer part b. (If I am not mistaken, $\begin{bmatrix}13\\4\end{bmatrix}$ is close to the answer.)
For part c, think about the definition of range, and the relationship between vectors in the range and $T$. It should consist of vectors in $\mathbb{R}^2$, by looking at $T$. Thus, you need at most 2 vectors (any more than that, and the set won't be linearly independent anymore).
Part d is true if you can write $\begin{bmatrix}\pi\\\sqrt{2}\end{bmatrix}$ as a linear combination of the two linearly independent vectors in part c, and false otherwise.
Best Answer
Hints:
a) Take the dot product of the vectors and it should be zero for any two different vectors.
b) Divide each vector by its length.
c) $v = \alpha b_1 + \beta b_2 + \gamma b_3 $
To determine the $\alpha,\beta$, and $\gamma$, take the dot product of $v$ with $b_1,b_2$, and $b_3$ and note that $b_i.||b_i||=1,\, \forall i=1..3$. Note that, $\alpha,\beta$, and $\gamma$ are known as the Fourier coefficients.