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A concise way to get both answers is to place the square in a convenient coordinate system and solve for the coordinates of E and F. A workable choice is A=(5,5),\quad B=(0,5),\quad C=(0,0),\quad D=(5,0), so that E lies on the left edge BC and F lies on the bottom edge CD. One may set E=(0,e),\quad F=(f,0), with 0<e<5 and 0<f<5. 1. Use \angle BEA=62^\circ to find E. The vectors \overrightarrow{EB} = B - E = (0,5)-(0,e) = (0,5-e), \quad \overrightarrow{EA} = A - E = (5,5)-(0,e) = (5,5-e), must make a 62^\circ angle. In particular, \[ \cos 62^\circ \;=\; \frac{\overrightarrow{EB}\cdot \overrightarrow{EA}}{\lVert EB\rVert\,\lVert EA\rVert} \frac{(5-e)^2}{(5-e)\sqrt{5^2+(5-e)^2}} \;=\; \frac{5-e}{\sqrt{25+(5-e)^2}}. \] Solving numerically gives 5-e\approx 2.66, hence e\approx 2.34. 2. Use \angle EAF=45^\circ to find F. Now \overrightarrow{AE}=(0-e,2.34-5) and \overrightarrow{AF}=(f-5,-5) must make a 45^\circ angle. Imposing \[ \cos 45^\circ \;=\; \frac{\overrightarrow{AE}\cdot \overrightarrow{AF}}{\lVert AE\rVert\,\lVert AF\rVert}, \] one finds f\approx 3.47. 3. Locate \angle 1 at F. From the diagram, \angle 1 is the angle at F between the segments FE and FD. In coordinates, \overrightarrow{FE} \;=\; E-F = (0,2.34)-(3.47,0)=( -3.47,\,2.34),\quad \overrightarrow{FD} \;=\; D-F = (5,0)-(3.47,0)=(1.53,\,0). A straightforward dot‐product computation shows \[ \cos(\angle 1) \;=\; \frac{\overrightarrow{FE}\cdot \overrightarrow{FD}} {\lVert FE\rVert\,\lVert FD\rVert} \;\approx\; -0.829, \] giving \angle 1 \;\approx\; 146^\circ. 4. Perimeter of \triangle CEF. The vertices are C=(0,0),\quad E=(0,2.34),\quad F=(3.47,0), so the side lengths are CE=2.34,\quad CF=3.47,\quad EF=\sqrt{(-3.47)^2 + 2.34^2}\,\approx\,4.19. Their sum is very close to 2.34 + 3.47 + 4.19 \approx 10. Hence the answers are \boxed{\angle 1 \approx 146^\circ \quad\text{and}\quad \text{perimeter of }\triangle CEF \approx 10\text{ cm.}}