Answer:
Part 1)
![SA=40\ units^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/y5anqldyudzwes9dksj12kfk0pya665sp0.png)
Part 2)
![SA=6.25\ units^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/3qnj725nckuhsyovq5449f2v5bis8ihrts.png)
Explanation:
Part 1) we know that
The surface area of a square pyramid is equal to the area of the square base plus the area of its four triangular faces
so
![SA=b^2+4[(1)/(2)(b)(h)]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/khc67vnomhwqq1z3zuib825phxexbr8yjl.png)
we have
![b=4\ units\\h=3\ units](https://img.qammunity.org/2021/formulas/mathematics/middle-school/92366qzwj3wmp3r150o8kblhqkjkj2fbiu.png)
substitute
![SA=4^2+4[(1)/(2)(4)(3)]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/seheaj7jf451pys23m267hp97f1j5zq3o0.png)
![SA=16+24=40\ units^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/qt1zwek0vg1b6if28ah3r1kqfhlinn36we.png)
Part 2) we know that
The surface are of the cube is equal to the area of its six square faces
so
![SA=6b^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/sl0x76sg6p7zbol9jmcddhjs7pqccotwdt.png)
we have
![b=2.5\ units](https://img.qammunity.org/2021/formulas/mathematics/middle-school/tyj973kjzcju7ouljc0osf0bodwn0k454o.png)
substitute
![SA=2.5^2\\SA=6.25\ units^2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ew2ht5pmfntwbmecj1ut5tlaxwlrcelwx4.png)