(a)

Without temperature change:

$\sigma = \dfrac{P}{A} = \dfrac{1200}{0.25} = 4800 \, \text{ psi }$
$\sigma = 4.8 \, \text{ksi} \lt 10 \, \text{ksi}$

A drop of temperature is needed to increase the stress to 10 ksi. See figure below.

$\delta = \delta_T + \delta_{st}$

$\dfrac{\sigma \, L}{E} = \alpha \, L \, (\Delta T) + \dfrac{PL}{AE}$

$\sigma = \alpha \, E \, (\Delta T) + \dfrac{P}{A}$

$10\,000 = (6.5 \times 10^{-6})(29 \times 10^6)(\Delta T) + \dfrac{1200}{0.25}$

$\Delta T = 27.59^\circ \text{F}$

Required temperature: *(temperature must drop from 40°F)*

$T = 40 - 27.59 = 12.41^\circ \text{F}$ *answer*

(b) Temperature at which the stress will be zero:

From the figure below:

$\delta = \delta_T$

$\dfrac{PL}{AE} = \alpha \, L \, (\Delta T)$

$P = \alpha AE(T_f \, - \, T_i)$

$1200 = (6.5 \times 10^{-6})(0.25) (29 \times 10^6)(T_f - 40)$

$T_f = 65.46^\circ\text{F}$ *answer*

shouldn't you have to increase the temperature to increase the tensile stress and decrease the temperature to decrease the tensile stress?

No. Temperature change itself won't induce stress. Stress will occur only if you restrain the expansion or contraction due to temperature change.