Answer
(a) greater than
(b) $3.4N$
(c) $0.41Kg$
Work Step by Step
(a) We know that the horizontal component of string 1 balances the large horizontal component of string 2. Thus, string 1 supports most of the weight of the picture and hence the tension in string 1 is greater than string 2.
(b) We can calculate the tension in string 1 as
$\Sigma F_x=-T_1cos\theta+T_2cos\theta=0$
This simplifies to:
$T_1=T_2 \frac{cos\theta_2}{cos\theta_1}$
We plug in the known values to obtain:
$T_1=(1.7)\frac{cos 32^{\circ}}{cos 65^{\circ}}$
$T_1=3.4N$
(c) We can find the mass of the picture as follows:
$\Sigma F_y=T_1sin\theta_1+T_2sin\theta_2-W=0$
$\implies W=T_1sin\theta+T_2sin\theta_2$
We plug in the known values to obtain:
$W=(3.4)sin65^{\circ}+(1.7)sin32^{\circ}$
$W=4.0N$
Now $m=\frac{W}{g}$
We plug in the known values to obtain:
$m=\frac{4.0}{9.81}$
$m=0.41Kg$