Answer :
(a) The heat released by the metal is -312.48 J
(b) The specific heat of the metal is 
Explanation :
<u>For part A :</u>
Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water
![q=[q_1+q_2]](https://tex.z-dn.net/?f=q%3D%5Bq_1%2Bq_2%5D)
![q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]](https://tex.z-dn.net/?f=q%3D%5Bc_1%5Ctimes%20%5CDelta%20T%2Bm_2%5Ctimes%20c_2%5Ctimes%20%5CDelta%20T%5D)
where,
q = heat released by the metal
= heat absorbed by the calorimeter
= heat absorbed by the water
= specific heat of calorimeter = 
= specific heat of water = 
= mass of water = 50.0 g
= change in temperature = 
Now put all the given values in the above formula, we get:
![q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]](https://tex.z-dn.net/?f=q%3D%5B%2851.5J%2F%5EoC%5Ctimes%201.2%5EoC%29%2B%2850.0g%5Ctimes%204.184J%2Fg%5EoC%5Ctimes%201.2%5EoC%29%5D)

Thus, the heat released by the metal is -312.48 J
<u>For part B :</u>

q = heat released by the metal = -312.48 J
m = mass of metal = 43.1 g
c = specific heat of metal = ?
= change in temperature = 
Now put all the given values in the above formula, we get:


Thus, the specific heat of the metal is 